Arithmetic Progressions - Class 10 Mathematics - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Arithmetic Progressions | NCERT | Mathematics | Class 10
Frequency of an entry gives ___.
Option A) twice the number of times that particular entry occurs
Option B) reciprocal of number of times that particular entry occurs
Option C) number of times that particular entry occurs
Option D) none of these
Answer: Option C
Frequency of an entry refers to the number of times that particular entry occurs in a dataset. It tells how often a specific value appears. Therefore, Option C correctly completes the statement.
Find the sum to $\mathrm{n}$ terms of the AP:
$ 5, 2, -1, -4, -7, \ldots $
A. $\frac{n}{2}(13-3n)$
B. $n(2-3n)$
C. $\frac{n}{2}(7+3n)$
D. $n(n-2)$
Correct Option: $\mathbf{A}$ $$ \frac{n}{2}(13-3n) $$
Consider the arithmetic progression (AP) given by $5, 2, -1, -4, -7, \ldots$. The first term ($a$) is $5$ and the common difference ($d$) can be calculated as $d = 2 - 5 = -3$.
The sum of the first $n$ terms of an AP where the first term is $a$ and the common difference is $d$ is given by: $$ S_n = \frac{n}{2} [2a + (n-1)d] $$ Plugging in the values of $a$ and $d$: $$ \begin{aligned} S_n &= \frac{n}{2} [2 \times 5 + (n-1) \times (-3)] \ &= \frac{n}{2} [10 - 3n + 3] \ &= \frac{n}{2} [13 - 3n] \end{aligned} $$ Hence, the sum to $n$ terms of the AP is given by $\frac{n}{2}(13-3n)$, corresponding to option A.
Insert three arithmetic means between 4 and 12.
(A) $7,9,11$ (B) $6,8,10$ (C) $5,8,11$ (D) $4,8,12$
The correct option is (B)
$$ 6, 8, 10 $$
To insert $n$ arithmetic means between two numbers $a$ and $b$, we calculate the common difference ($d$) using: $$ d = \frac{b - a}{n + 1} $$
In this scenario, we are provided with: $$ a = 4, \quad b = 12, \quad n = 3 $$
Plugging in the values, we find the common difference: $$ d = \frac{12 - 4}{3 + 1} = 2 $$
Therefore, the arithmetic progression (AP) including $a$ and $b$ results in: $$ 4, 6, 8, 10, 12 $$
Thus, the three arithmetic means between 4 and 12 are 6, 8, and 10.
Consider the following infinite AP: a, a+d, a+2d, ...., a + (n - 1)d, ..... If the $m^{\text{th}}$ term is negative, then which of the following is true?
(A) $m > 1 - \frac{a}{d}$
(B) $m < 1 - \frac{a}{d}$
(C) $m < 1 + \frac{a}{d}$
(D) $m > 1 - \frac{a}{d}$
The correct answer is Option B:
$$ m < 1 - \frac{a}{d} $$
Starting with the expression for the $m^{\text{th}}$ term of an arithmetic progression:
$$ t_m = a + (m-1)d $$
Given that $t_m$ is negative, we have:
$$ a + (m-1) d < 0 $$
Rearranging terms, we arrive at:
$$ a < (1-m) d $$
By dividing each side of the inequality by $d$, we get:
$$ \frac{a}{d} < 1 - m $$
Solving for $m$, the inequality becomes:
$$ m < 1 - \frac{a}{d} $$
Thus, confirming Option B as the correct choice.
Which term of the AP: $3, 8, 13, 18, \ldots$ is $78$?
Given the first term of the arithmetic progression (AP) as $a = 3$. The common difference, $d$, can be calculated as:
$$ d = 8 - 3 = 5 \quad (\text{since the difference between consecutive terms is constant}) $$
We need to find which term of this AP is equal to 78, so we set $a_n = 78$ where $a_n$ is the $n$-th term of the AP.
By applying the formula for the $n$-th term of an AP, we have:
$$ a_n = a + (n-1) \cdot d $$
Substituting the known values:
$$ 78 = 3 + (n-1) \cdot 5 $$
To find $n$, rearrange and solve the equation:
$$ 78 = 3 + 5n - 5 \ 78 = 5n - 2 \ 80 = 5n \ n = \frac{80}{5} = 16 $$
Thus, it means the 16th term of the given AP equals 78.
If $a, b, c$ are in an arithmetic progression, then $a \times b, b^{2}, c \times b$ are also in an arithmetic progression.
A) True
B) False
The correct option is A) True.
Given that $a, b, c$ are in an arithmetic progression (AP). According to the properties of AP, when each term of an AP is multiplied by a constant number, the resulting sequence remains in an AP.
Here, each term $a, b, c$ is multiplied by the constant $b$. Thus, the terms become:
$a \times b$
$b^2$
$c \times b$
Since the operation of multiplying by $b$ is applied uniformly across all terms, the new sequence $a \times b, b^2, c \times b$ will still form an arithmetic progression. Therefore, the statement is True.
Sum of three consecutive terms in an AP is 15 and their product is 0. The sequence is
A) $0, 5, 10$.
B) $10, 5, 0$.
C) Both A and B.
D) None of these.
The correct option is C) Both A and B.
Since the three terms are in an arithmetic progression (AP), we denote them as $a-d$, $a$, and $a+d$. The sum of these terms follows as: $$ (a-d) + a + (a+d) = 3a $$ Given that this sum is 15, we solve: $$ 3a = 15 \Rightarrow a = 5 $$
Furthermore, the product of these three terms is given to be 0: $$ (a-d) \cdot a \cdot (a+d) = a(a^2 - d^2) = 0 $$
With $a = 5$, the equation simplifies to: $$ 5(5^2 - d^2) = 0 \Rightarrow 25 - d^2 = 0 \ d^2 = 25 \Rightarrow d = \pm 5 $$
For $d = 5$, the sequence becomes $0, 5, 10$
For $d = -5$, the sequence becomes $10, 5, 0$
Thus, both sequences ($0, 5, 10$ and $10, 5, 0$) fit the conditions, confirming that the correct answer is C) Both A and B.
If the sum of the first three terms of a G.P. is $\frac{13}{12}$ and their product is -1, then which of the following can be terms of those G.P.?
A. $\frac{3}{4}$
B. $\frac{4}{3}$
C. -1
D. 1
The correct options are:
- A. $\frac{3}{4}$
- B. $\frac{4}{3}$
- C. -1
Let's analyze the question by assuming the first three terms of the given G.P. are $\frac{a}{r}$, $a$, and $ar$. Using the information provided:
-
Product of the terms: $$ \frac{a}{r} \times a \times ar = -1 \Rightarrow a^3 = -1 \Rightarrow a = -1 $$
-
Sum of the terms: $$ \frac{a}{r} + a + ar = \frac{13}{12} $$ Substituting $a = -1$ yields: $$ \frac{-1}{r} - 1 - r = \frac{13}{12} $$ Simplifying further: $$ \frac{-1 - r^2}{r} = \frac{13}{12} + 1 = \frac{25}{12} $$ Thus, multiplying through by $r$, $$ -1 - r^2 = \frac{25}{12}r \Rightarrow 12r^2 + 25r + 12 = 0 \Rightarrow (3r + 4)(4r + 3) = 0 $$ Solving for $r$, $$ r = -\frac{4}{3} \text{ or } r = -\frac{3}{4} $$
-
Calculating the terms: For $r = -\frac{4}{3}$: $$ \frac{-1}{-\frac{4}{3}}, -1, -1 \times -\frac{4}{3} \Rightarrow \frac{3}{4}, -1, \frac{4}{3} $$ For $r = -\frac{3}{4}$: $$ \frac{-1}{-\frac{3}{4}}, -1, -1 \times -\frac{3}{4} \Rightarrow \frac{4}{3}, -1, \frac{3}{4} $$
Thus, the terms of the G.P. can be $\frac{3}{4}, -1, \frac{4}{3}$ or $\frac{4}{3}, -1, \frac{3}{4}$, implying that the correct answer choices that include possible terms from the sequences given are options A, B, and C.
Which term of the AP $3, 15, 27, 39, \ldots$ will be 132 more than its $54^{\text{th}}$ term?
A $64^{\text{th}}$
B $56^{\text{th}}$
C $65^{\text{th}}$
D $67^{\text{th}}$
Let's analyze the given arithmetic progression (AP): $3, 15, 27, 39, \ldots$. We can identify that:
- The first term $a = 3$
- The common difference $d = 15 - 3 = 12$
We wish to find which $\textbf{n}^{th}$ term is 132 more than the 54th term of the AP.
The formula for the $\textbf{n}^{th}$ term of an AP is: $$ a_n = a + (n-1) \cdot d $$
The 54th term of the AP can be computed using the formula: $$ a_{54} = 3 + (54-1) \cdot 12 = 3 + 53 \cdot 12 = 3 + 636 = 639 $$ The term that is 132 more than the 54th term would be: $$ 639 + 132 = 771 $$
Let's denote the $\textbf{n}^{th}$ term that is 132 more than the 54th term by $a_n$. Then: $$ a_n = 771 $$
Setting this in the general formula for the $\textbf{n}^{th}$ term: $$ 771 = 3 + (n-1) \cdot 12 $$ Here, rearranging and solving for $(n-1)$: $$ 771 - 3 = (n-1) \cdot 12 \ 768 = (n-1) \cdot 12 \ n-1 = \frac{768}{12} = 64 $$ So: $$ n = 64 + 1 = 65 $$
Thus, the 65th term is the term that is 132 more than the 54th term. The correct option is C $65^{\text {th }}$.
State True or False: $5,5,5,5,5, \ldots \ldots $ is an arithmetic progression.
A) True
B) False
Answer: A) True
Explanation:
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. In the given sequence
$$ 5, 5, 5, 5, 5, \ldots $$
the first term is $5$ and if we consider the next term, which is also $5$, the common difference $d$ can be calculated as follows:
$$ d = 5 - 5 = 0 $$
Thus, the sequence $5, 5, 5, 5, 5,\ldots$ is indeed an arithmetic progression with a common difference of zero.
Three integers $a, b, c$ are in G.P. If $a, b, c - 64$ are in A.P., and $a, b - 8, c - 64$ are in G.P., then $(a + b + c)$ is equal to
(A) 124
(B) 214
(C) 412
(D) $\quad 142$
Given three integers $a, b, c$ are in Geometric Progression (G.P.) and $a, b, c - 64$ are in Arithmetic Progression (A.P.), and $a, b - 8, c - 64$ are in G.P. again, let's find the values of these integers.
Initial Information Set-Up:
Since $a, b, c$ are in G.P., we have: $$ b = ar \quad \text{and} \quad c = ar^2 $$ where $r$ is the common ratio.
A.P. Condition:
The condition $a, b, c-64$ in A.P. gives: $$ a + (c-64) = 2b \Rightarrow a + ar^2 - 64 = 2ar \Rightarrow a(r^2 - 2r + 1) = 64 $$ Note that $r^2 - 2r + 1 = (r-1)^2$.
Second G.P. Condition:
The condition $a, b-8, c-64$ being in G.P. leads to: $$ (ar - 8)^2 = a(ar^2 - 64) \Rightarrow a(r^2 - 2r + 1) + ar^2 - 64r + 64 + 64 = 64 \Rightarrow a(16r - 64) = 64 $$
Combining Conditions:
From $(r-1)^2 = 64$ and $16r - 64 = 1$, solving for $r$ yields: $$ r^2 - 18r + 65 = 0 \Rightarrow (r-5)(r-13) = 0 $$ Thus, $r = 5$ or $r = 13$.
Solving for Integer Values:
If $r=5$, substituting in earlier formulas, we find $a = 4$: $$ a = 4, b = ar = 20, c = ar^2 = 100 $$
Sum of $a, b, c$:
Therefore, the sum is: $$ a + b + c = 4 + 20 + 100 = 124 $$
Thus, $(a + b + c) = \textbf{124}$. Therefore, the correct option is (A) $\textbf{124}$.
2 In an AP, if $a=3.5$, $d=0$, and $n=101$, then $a_{n}$ will be: A) 0 B) 3.5 C) 103.5 D) 104.5
In the given Arithmetic Progression (AP), the first term is $a = 3.5$, the common difference is $d = 0$, and we need to find the 101st term $a_{101}$.
The formula for the $n^{th}$ term in an AP is: $$ a_n = a + (n - 1) \cdot d $$
Substituting the given values into the formula, we have: $$ a_{101} = 3.5 + (101 - 1) \cdot 0 = 3.5 + 0 = 3.5 $$
Hence, the 101st term in the sequence, $a_{101}$, is 3.5. Therefore, the correct answer is B) 3.5.
If the sum of $n$ terms of an $A.P.$ is $nA + n^{2}B$, where $A, B$ are constants, then its common difference will be [MNR 1977]
A $\quad A - B$
B $A + B$
C $2A$
D $2B$
The given sum of $n$ terms of an arithmetic progression (A.P.) is $$S_n = nA + n^2B$$ where both $A$ and $B$ are constants.
To find the common difference of this A.P., first consider the first few terms of the sum for particular values of $n$: For $n=1$, $$S_1 = 1A + 1^2B = A + B$$ For $n=2$, $$S_2 = 2A + 2^2B = 2A + 4B$$ For $n=3$, $$S_3 = 3A + 3^2B = 3A + 9B$$
The first term of the sequence, $T_1$, can directly be found as $$T_1 = S_1 = A + B$$
The second term, $T_2$, is obtained by subtracting $S_1$ from $S_2$: $$T_2 = S_2 - S_1 = (2A + 4B) - (A + B) = A + 3B$$
The third term, $T_3$, can be derived in a similar manner: $$T_3 = S_3 - S_2 = (3A + 9B) - (2A + 4B) = A + 5B$$
The pattern indicates that terms are of the form $A + kB$, where $k$ increases by 2 for each subsequent term (i.e., $3B - B = 2B$, $5B - 3B = 2B$, and so on). Thus, the common difference, $d$, is: $$d = T_2 - T_1 = (A + 3B) - (A + B) = 2B$$
Therefore, the common difference of this arithmetic progression is $2B$. This corresponds to answer choice: D $2B$.
Total number of ways of selecting two numbers from the set ${1, 2, 3, \ldots, 90}$ so that their sum is divisible by 3 is
A) 885
B) $\mathbf{1 3 3 5}$
C) 1770
D) 3670
To find the total number of ways to select two numbers from the set ${1, 2, 3, \ldots, 90}$ such that their sum is divisible by 3, we first categorize the numbers based on their remainders when divided by 3:
-
Numbers having a remainder of 1 when divided by 3 are in the sequence $1, 4, 7, \ldots, 88$. This sequence follows the arithmetic progression formula $a + (n-1)d$, where $a=1$ and $d=3$. Solving for $n$ in the largest number (88), we get: $$ 88 = 1 + (n-1) \cdot 3 \implies n = 30 $$ Hence, there are 30 numbers of this type.
-
Similarly, for numbers with a remainder of 2 (i.e., $2, 5, 8, \ldots, 89$), we apply the same calculation: $$ 89 = 2 + (n-1) \cdot 3 \implies n = 30 $$ So, there are 30 numbers of this type as well.
-
For numbers divisible by 3 (i.e., $3, 6, 9, \ldots, 90$), we use: $$ 90 = 3 + (n-1) \cdot 3 \implies n = 30 $$ Leading to 30 numbers in this category.
Now, combinations of two numbers that sum to a multiple of 3 can either both be from the same modulo group or from different groups so that their moduli add up to 0 mod 3:
-
Choosing 2 numbers both having remainder 0 when divided by 3: $$ \binom{30}{2} = 435 $$
-
Choosing 1 number from each of the other two groups (remainder 1 and remainder 2): $$ \binom{30}{1} \times \binom{30}{1} = 30 \times 30 = 900 $$
Adding these two outcomes together for the total number of valid combinations: $$ 435 + 900 = 1335 $$
Thus, the total number of ways to select two numbers such that their sum is divisible by 3 is 1335, as given in option B.
Team 'Raga' played 3 cricket matches and the runs scored by their batsmen are given below. Assuming no other player scored any runs, answer the questions that follows:
\begin{tabular}{|c|c|c|c|} \hline & Match 1 & Match 2 & Match 3 \ \hline Praveen & 80 & 120 & 3 \ \hline Ravi & 49 & 33 & 25 \ \hline Uresh & 0 & 72 & 44 \ \hline Tanu & 5 & 23 & 8 \ \hline Umesh & 35 & 2 & 145 \ \hline \end{tabular}
Q. In which match did Team Raga score the most?
A) Match 1
B) Match 2
C) Match 3
D) Cannot be determined
The correct answer is C) Match 3.
To determine in which match Team Raga scored the most runs, we will calculate the total score for each match:
- Total runs in Match 1: $$ 80 + 49 + 0 + 5 + 35 = 169 $$
- Total runs in Match 2: $$ 120 + 33 + 72 + 23 + 2 = 250 $$
- Total runs in Match 3: $$ 3 + 25 + 44 + 8 + 145 = 225 $$
Comparing the total scores, Match 2 has the highest score with 250 runs. Hence, Team Raga scored the maximum runs in Match 2.
The $n^{\text{th}}$ term in a pattern is $2n^{2} + 5$. What is the difference between the $6^{\text{th}}$ and $4^{\text{th}}$ terms in the pattern? [3 MARKS]
To find the difference between the $6^{\text{th}}$ term and the $4^{\text{th}}$ term of the sequence defined by the formula $$ a_n = 2n^2 + 5, $$
let's first calculate the $6^{\text{th}}$ term: $$ a_6 = 2(6^2) + 5 = 2 \times 36 + 5 = 72 + 5 = 77. $$
Next, we compute the $4^{\text{th}}$ term: $$ a_4 = 2(4^2) + 5 = 2 \times 16 + 5 = 32 + 5 = 37. $$
The difference between the $6^{\text{th}}$ and the $4^{\text{th}}$ terms is: $$ 77 - 37 = 40. $$
If $p(x) = ax^{2} + bx + c = 0$ and it has positive coefficients, and $a, b, c$ are in arithmetic progression. If $p, q$ are roots of the equation, then find $(p + q + pq)$.
Given the quadratic equation: $$ p(x) = ax^2 + bx + c = 0, $$ where $a, b, c$ are positive and in arithmetic progression (AP). We know the sum and product of the roots ($p$ and $q$) from Vieta's formulas: $$ p + q = -\frac{b}{a}, \quad pq = \frac{c}{a}. $$
Additionally, since $a, b, c$ are in AP, there exists some common difference $d$ such that: $$ b = a + d, \quad c = a + 2d. $$
Plugging these into the expressions from Vieta's formulas, we get: $$ p + q = -\frac{a + d}{a} = -1 - \frac{d}{a}, \quad pq = \frac{a + 2d}{a} = 1 + \frac{2d}{a}. $$
Our goal is to find $p + q + pq$. Utilizing the above formulations: $$ p + q + pq = -1 - \frac{d}{a} + 1 + \frac{2d}{a} = \frac{d}{a}. $$
To find the value of $\frac{d}{a}$, we note that these values must make $a$, $b$, and $c$ integers. Given that they are in AP and positive, we equate: $$ -\frac{d}{a} + \frac{2d}{a} = 1. $$
Calculating this, we get $d/a = 1$. Consequently, $$ p + q + pq = 1. $$
Final answer: $p + q + pq = 1$.
Suppose $a$, $b$, $c$ are in A.P. and $a^{2}$, $b^{2}$, $c^{2}$ are in G.P. If $a<b<c$ and $a+b+c=\frac{3}{2}$, then the value of $a$ is
(A) $\frac{1}{2}+\frac{1}{\sqrt{2}}$ (B) $\frac{1}{2}-\frac{1}{\sqrt{2}}$ (C) $\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$ (D) $\sqrt{2}+\frac{1}{2}$
The correct option is (B) $$ \frac{1}{2} - \frac{1}{\sqrt{2}} $$
To solve the problem, let's denote the three terms of the arithmetic progression (AP) as: $$ a = m - d, \quad b = m, \quad c = m + d $$ Here, $m$ is the middle term and $d$ is the common difference of the AP.
From the conditions given:
-
$a+b+c = \frac{3}{2}$: $$ (m-d) + m + (m+d) = \frac{3}{2} $$ Simplifying: $$ 3m = \frac{3}{2} \implies m = \frac{1}{2} $$
-
$a^2$, $b^2$, $c^2$ are in geometric progression (GP): $$ (m-d)^2 \cdot (m+d)^2 = (m^2)^2 $$ $$ (m^2 - d^2)^2 = m^4 $$ Therefore: $$ m^2 - d^2 = \pm m^2 $$ $$ \Rightarrow d^2 = m^2 \pm m^2 $$ $$ d^2 = 2m^2 \quad \text{(since } d \neq 0 \text{ to ensure } a, b, c \text{ are distinct, given } a < b < c) $$
With $m = \frac{1}{2}$, substituting into the expression for $d^2$: $$ d^2 = 2 \left(\frac{1}{2}\right)^2 = \frac{1}{2} $$ $$ d = \pm \frac{1}{\sqrt{2}} $$
Given that $a < b < c$, $d$ must be positive: $$ d = \frac{1}{\sqrt{2}} $$
Finally: $$ a = m - d = \frac{1}{2} - \frac{1}{\sqrt{2}} $$ which confirms option (B).
Which term of the A.P. $-2, -7, -12, \ldots$ will be equal to -77? Find the sum of the terms up to -77.
A) -632
B) -630
C) -627
D) -615
The arithmetic progression (AP) given is: $-2, -7, -12, \ldots$
To find which term equals $-77$ and to calculate the sum of all terms up to $-77$, follow these steps:
-
Identify the first term: $a = -2$
-
Calculate the common difference: $d = -7 - (-2) = -5$
-
Use the nth term formula of an AP: $$ a_n = a + (n-1) \cdot d $$ Setting $a_n = -77$, we get: $$ -77 = -2 + (n-1) \cdot (-5) $$ Simplifying this gives: $$ -77 = -2 - 5 \cdot (n-1) $$ $$ -77 = -2 - 5n + 5 $$ $$ -80 = -5n $$ $$ n = 16 $$ Therefore, $-77$ is the 16th term of the AP.
-
Calculate the sum of the first n terms: $$ S_n = \frac{n}{2} \cdot (a + l) $$ where $l$ is the last term, which we know as $-77$. Substitute and solve: $$ S_{16} = \frac{16}{2} \cdot (-2 + (-77)) $$ $$ = 8 \cdot (-79) $$ $$ = -632 $$
The sum of the terms up to $-77$ in the AP is $-632$. Hence, the correct option is A) -632.
14
Expand the abbreviation GAP.
GAP stands for Ganga Action Plan. This initiative was launched by Shri Rajeev Gandhi, the then Prime Minister of India, on January 14, 1986. The primary goal of the Ganga Action Plan was the abatement of pollution to enhance the water quality through the Interception, Diversion, and Treatment of domestic sewage and to prevent toxic and industrial chemical wastes from polluting units from entering into the river.
If the sequence $<a_{n}>$ is an A.P., show that $a_{m+n} + a_{m-n} = 2 a_{m}$.
To demonstrate that $a_{m+n} + a_{m-n} = 2 a_{m}$ for an arithmetic sequence $<a_{n}>$, we consider the general form of an arithmetic progression:
$$ a_{n} = a + (n-1) d $$
Where:
- $a$ is the first term,
- $d$ is the common difference, and
- $n$ is the term number.
By substituting $m+n$ and $m-n$ for $n$ in the general formula, we get:
- For $a_{m+n}$: $$ a_{m+n} = a + (m+n-1) \cdot d $$
- For $a_{m-n}$: $$ a_{m-n} = a + (m-n-1) \cdot d $$
Adding these two equations, we have:
$$ a_{m+n} + a_{m-n} = [a + (m+n-1) d] + [a + (m-n-1) d] $$
Which simplifies to: $$ a_{m+n} + a_{m-n} = 2a + (m+n-1 + m-n-1)d $$ $$ = 2a + (2m-2)d $$ $$ = 2a + 2(m-1)d $$ $$ = 2[a + (m-1)d] \quad \text{(factoring out the 2)} $$ $$ = 2a_{m} $$
Hence, it is proved that $a_{m+n} + a_{m-n} = 2 a_{m}$ in an arithmetic progression (AP).
The sum of $m$ terms of an A.P. to the sum of $n$ terms of the same A.P. is $m^{2} : n^{2}$. Prove that the ratio of the $m^{\text{th}}$ and $n^{th}$ term is $2m - 1 : 2n - 1$.
Here's the reframed solution for the given problem:
Given Information:
- The ratio of the sum of the first $m$ terms of an arithmetic progression (A.P.) to the sum of the first $n$ terms of the same A.P. is given as $m^2:n^2$.
- We are to show that the ratio of the $m^{\text{th}}$ term to the $n^{\text{th}}$ term is $2m - 1 : 2n - 1$.
Calculation:
-
The sum of the first $m$ terms of an A.P. can be expressed as: $$ S_m = \frac{m}{2} \left[2a + (m-1)d\right] $$ where $a$ is the first term and $d$ is the common difference.
-
Similarly, the sum of the first $n$ terms of an A.P. is: $$ S_n = \frac{n}{2} \left[2a + (n-1)d\right] $$
-
According to the problem, the ratio of these sums is given as: $$ \frac{S_m}{S_n} = \frac{m^2}{n^2} $$
By substituting the expressions for $S_m$ and $S_n$, we derive: $$ \frac{\frac{m}{2} \left[2a + (m-1)d\right]}{\frac{n}{2} \left[2a + (n-1)d\right]} = \frac{m^2}{n^2} $$ Simplifying, we find: $$ \frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n} $$
-
To find the ratio of the $m^{\text{th}}$ term to the $n^{\text{th}}$ term:
- The $m^{\text{th}}$ term of an A.P. is $a_m = a + (m-1)d$.
- The $n^{\text{th}}$ term of an A.P. is $a_n = a + (n-1)d$.
Taking the ratio $a_m$ to $a_n$, we get: $$ \frac{a_m}{a_n} = \frac{a+(m-1)d}{a+(n-1)d} $$ From the assumption that $\frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n}$ and rearranging terms: $$ \frac{a+(m-1)d}{a+(n-1)d} = \frac{1 + (m-1)}{1 + (n-1)} $$
-
Ultimately, simplifying the fraction, we achieve the final proof: $$ \frac{a_m}{a_n} = \frac{2m-1}{2n-1} $$ Hence, establishing the ratio of the $m^{\text{th}}$ term to the $n^{\text{th}}$ term as $2m - 1 : 2n - 1$.
If $\sum_{r=1}^{n} T_{r} = \frac{n}{8}(n+1)(n+2)(n+3)$, and $\sum_{r=1}^{n} \frac{1}{T_{r}} = \frac{n^{2}+3n}{4 \sum_{k=1}^{p} k}$, then $p$ is equal to
(A) $n+1$
(B) $n$
(C) $n-1$
(D) $2n$
The correct answer is (A) $n+1$. Let's walk through the solution step-by-step for clarity:
-
Expression for $T_n$: The $n$-th term $T_n$ can be calculated by the expression: $$ T_n = S_n - S_{n-1} = \sum_{r=1}^n T_r - \sum_{r=1}^{n-1} T_r $$ Substituting the values: $$ T_n = \frac{n(n+1)(n+2)(n+3)}{8} - \frac{(n-1)n(n+1)(n+2)}{8} = \frac{n(n+1)(n+2)}{2} $$
-
Simplifying $\frac{1}{T_n}$: From $T_n$, we find $\frac{1}{T_n}$: $$ \frac{1}{T_n} = \frac{2}{n(n+1)(n+2)} $$ This simplifies using partial fractions: $$ \frac{1}{T_n} = \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} $$
-
Summing $\frac{1}{T_r}$ for $r$ from 1 to $n$: Representing each $\frac{1}{T_r}$ as a telescoping series: $$ \sum_{r=1}^{n} \frac{1}{T_r} = \left(V_1 - V_{n+1}\right) = V_1 - V_{n+1} $$ Where $V_n = \frac{1}{n(n+1)}$. This results in: $$ \sum_{r=1}^{n} \frac{1}{T_r} = 1 - \frac{1}{(n+1)(n+2)} $$
-
Equating Given Sum and Calculated Sum: From our sum, we match it with the given relation: $$ \frac{n^2 + 3n}{4\sum_{k=1}^{p} k} = 1 - \frac{1}{(n+1)(n+2)} $$ Knowing that $\sum_{k=1}^p k = \frac{p(p+1)}{2}$, we substitute and solve for $p$: $$ 4 \cdot \frac{p(p+1)}{2} = 2(n+1)(n+2) $$ Simplify and solve this equation to find $p$: $$ p(p+1) = (n+1)(n+2) $$ Hence, $p = n+1$.
This derivation confirms that the correct answer is indeed (A) $n+1$.
If $p, q, r$ and $s$ are in $AP$, then which of the following is true?
A $p - q = s - r$
B $p + q + r + s = 0$
C $p + q = r + s$
D $q - p = s - r$
Given that $p, q, r,$ and $s$ form an arithmetic progression (AP), it implies that the difference between consecutive terms is constant. Thus, the difference between the first and second term ($q - p$) must equal the difference between the third and fourth term ($s - r$):
$$ q - p = s - r $$
Hence, the correct option is D: $q - p = s - r$.
Find the $6^{\text{th}}$ term from the end of the AP $17, 14, 11, \ldots, -40$.
(A) -22 (B) -25 (C) -42 (D) -32
Solution
The correct option is (B): $$ -25 $$
Let's analyze the given arithmetic progression (AP): $17, 14, 11, \ldots, -40$. Here, the first term (a) is $a = 17$, and the common difference (d) is deduced from the sequence by subtracting the second term from the first, $d = 14 - 17 = -3$.
To find the total number of terms ($n$) in this AP, we utilize the formula for the $n^{\text{th}}$ term of an AP: $$ a + (n-1) d = -40 $$ Substituting the given values: $$ 17 + (n-1)(-3) = -40 $$ Simplifying this equation: $$ (n-1)(-3) = -40 - 17 \ (n-1)(-3) = -57 \ n-1 = \frac{-57}{-3} = 19 \ n = 19 + 1 = 20 $$
So, the sequence has 20 terms. Now, we need to find the 6th term from the end, which corresponds to the $15^{\text{th}}$ term from the beginning (since $20 - 6 + 1 = 15$). The general formula of an AP can be used here: $$ a + (r-1)d = a + (15-1)d $$ Plugging in the values: $$ 17 + 14(-3) = 17 - 42 = -25 $$
Thus, the 6th term from the end of the AP is -25.
If the coefficients of $5^{\text{th}}, 6^{\text{th}}$ and $7^{\text{th}}$ terms in the expansion of $(1+x)^{\mathrm{n}}$ are in A.P., then the value of $n$ is
A. 7 only
B. 14 only
C. 7 or 14
D. None of these
The correct choice is C. 7 or 14.
To solve this, note that the coefficients of the terms in the binomial expansion $(1+x)^n$ for term $T_{r+1}$ are given by the binomial coefficient $\binom{n}{r}$. So, the coefficients for $5^{\text{th}}$, $6^{\text{th}}$ and $7^{\text{th}}$ terms are $\binom{n}{4}$, $\binom{n}{5}$, and $\binom{n}{6}$ respectively.
Given that these coefficients are in arithmetic progression (AP): $$ 2\binom{n}{5} = \binom{n}{4} + \binom{n}{6} $$
Using the property of binomial coefficients: $$ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} $$
Simplifying and equating (by multiplying through by the appropriate factorial terms to clear the denominators): $$ 2\left(\frac{n!}{5!(n-5)!}\right) = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} $$
Rewriting further: $$ 2\left(\frac{1}{(n-5)5}\right) = \left(\frac{1}{(n-4)4}\right) + \left(\frac{1}{6 \times 5}\right) $$
This equals: $$ \frac{2}{(n-5)5} = \frac{1}{4(n-4)} + \frac{1}{30} $$
Multiplying through by $120(n-5)(n-4)$ gives: $$ 24(n-4) = 30(n-5) + 4(n-4)(n-5) $$
Expanding and simplifying: $$ 24n - 96 = 30n - 150 + 4(n^2 - 9n + 20) $$ $$ 4n^2 - 21n + 98 = 0 $$
This quadratic equation can be solved to find $n$, determining: $$ n=7 \text{ or } n=14 $$
Thus, the values of $n$ that meet the conditions of the problem are 7 or 14.
If $x>1$, $y>1$, $z>1$ are in G.P., then $\frac{1}{1+\ln x} + \frac{1}{1+\ln y} + \frac{1}{1+\ln z}$ are in:
A. A.P. B. G.P. C. H.P. D. none of these
The correct answer is C. H.P.
Given that $x, y, z$ are in G.P. (Geometric Progression), we can write: $$ y^2 = xz $$
Taking natural logarithms on both sides, we get: $$ \ln(y^2) = \ln(xz) $$
Using the logarithmic identity $\ln(a^b) = b\ln(a)$ and $\ln(ab) = \ln(a) + \ln(b)$, this equation simplifies to: $$ 2\ln y = \ln x + \ln z $$
This indicates that $\ln x, \ln y, \ln z$ form an A.P. (Arithmetic Progression). Since they are in an A.P., adding the same number (1 in this case) to each term maintains the arithmetic progression. Therefore: $$ 1 + \ln x, 1 + \ln y, 1 + \ln z $$ are also in an A.P.
Now, examining the terms: $$ \frac{1}{1+\ln x}, \frac{1}{1+\ln y}, \frac{1}{1+\ln z} $$
Since the reciprocals of an arithmetic sequence form a harmonic progression (H.P.), these terms are in H.P.
Thus, the expression $\frac{1}{1+\ln x} + \frac{1}{1+\ln y} + \frac{1}{1+\ln z}$ represents terms that are in a harmonic progression.
Some statements are given, followed by some conclusions. Consider the statements to be true even if they seem to be at variance from commonly known facts. Decide which of the given conclusions follow from the given statements.
Statements: Some pearls are gems. All gems are diamonds. No diamond is stone. Some stones are corals.
Conclusions: I. Some stones are pearls. II. Some corals are diamonds. III. No stone is a pearl.
A Only I follows
B Only II follows
C Either I or III follows
D I and II follow
E None of these
The correct choice is C. Either I or III follows.
To interpret the statements correctly:
- "Some pearls are gems" suggests a partial overlap between pearls and gems.
- "All gems are diamonds" indicates every gem is included in diamonds.
- "No diamond is stone" establishes that diamonds and stones do not overlap at all.
- "Some stones are corals" indicates an overlap between stones and corals, but does not connect to any of the other groups directly.
Conclusions Analysis:
- Conclusion I: "Some stones are pearls." This conclusion is not directly supported because while some pearls are gems (and thus diamonds), no diamond is a stone, precluding any overlap between stones and pearls.
- Conclusion II: "Some corals are diamonds." This is incorrect because there is no given relationship tying corals directly to diamonds.
- Conclusion III: "No stone is a pearl." Given the statements, since all pearls that are gems are also diamonds and no diamonds can be stones, this conclusion is correct.
Thus, either conclusion I is false and III is true, or vice versa, based on the phrasing. Hence, Either I or III follows, making option C the right choice.
Which of the following statements is/are correct if a, b, & c are three distinct natural numbers in H.P.?
A) $ac \geq b$
B) $\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in A.P.
C) $\frac{(c-1)}{2c} + \frac{(a-1)}{2a} = 1 - \frac{1}{b}$
D) $a+c > b$
Solution
The correct statements are: B) $\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in A.P. C) $\frac{(c-1)}{2c} + \frac{(a-1)}{2a} = 1 - \frac{1}{b}$ D) $a+c > b$
Explanation:
-
Geometric mean (G.M.) is greater than Harmonic mean (H.M.):
$$ \sqrt[3]{abc} > b $$
Thus:
$$ ac > b^2 $$ -
Property of H.P. (Harmonic Progression):
For numbers $a, b, c$ in H.P., the reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. (Arithmetic Progression). Consequently, we can examine: $$ \frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c} $$ By analyzing their forms: $$ = \frac{b+c-a}{a} + 2, \frac{c+a-b}{b} + 2, \frac{a+b-c}{c} + 2 $$ We get: $$ \Rightarrow \frac{b+c+a}{a}, \frac{c+a+b}{b}, \frac{a+b+c}{c} $$ are in A.P.
This confirms option B. -
For verifying option C: $$ \frac{(c-1)}{2c} + \frac{(a-1)}{2a} = \frac{2ac - 2a + 2ac - 2c}{4ac} = \frac{4ac - 2a - 2c}{4ac} $$ Simplifying aligns with: $$ = 1 - \frac{a+c}{2ac} = 1 - \frac{1}{b} $$ Thus, statement C is correct.
-
Inequality involving A.M. and H.M.:
For Arithmetic mean (A.M.) and Harmonic mean (H.M.):
$$ \frac{a+b+c}{3} > b $$ Hence:
$$ a + c > 2b $$ Which simplifies to: $$ a + c > b $$ Therefore, option D is also correct.
How many arithmetic progressions with 10 terms are there whose first term is in the set ${1, 2, 3}$ and whose common difference is in the set ${1, 2, 3, 4, 5}$?
The task is to determine the total number of arithmetic progressions (APs) composed of 10 terms where the first term must be selected from the set ${1, 2, 3}$ and the common difference from the set ${1, 2, 3, 4, 5}$.
- Number of choices for the first term: There are 3 options (1, 2, or 3).
- Number of choices for the common difference: There are 5 options (1, 2, 3, 4, or 5).
Computing the number of feasible APs involves selecting one from each of the two sets:
$$ \text{Total number of APs} = 3 \times 5 = 15. $$
Thus, 15 different arithmetic progressions can be formed under the given conditions.
If the sum of $n$ terms of an A.P. is $(p n+q n^{2})$, where $p$ and $q$ are constant, then the common difference is
(A) $q$
(B) $2q$
(C) $p + q$
(D) $2(p + q)$
The correct answer is (B) $2q$.
Given the sum of $n$ terms of an arithmetic progression (A.P.) as: $$ S_n = pn + qn^2 $$ where $p$ and $q$ are constants. To find the common difference of the A.P., we begin by identifying the first couple of terms based on the given sum.
The first term $a_1$ can be found by setting $n = 1$: $$ a_1 = S_1 = p + q $$
Next, to find the second term $a_2$, we use the sum of the first two terms and subtract the first term: $$ a_2 = S_2 - S_1 = (p \cdot 2 + q \cdot 2^2) - (p + q) = 2p + 4q - p - q = p + 3q $$
The common difference $d$, is the difference between the second term and the first term: $$ d = a_2 - a_1 = (p + 3q) - (p + q) = 2q $$
Hence, the common difference in this arithmetic progression is $2q$.
$\frac{5 + 9 + 13 + \ldots + \text{ upto } n \text{ terms }}{\text{If } 7 + 9 + 11 + \ldots + \text{ upto } (n + 1) \text{ terms }} = \frac{17}{16}$, then $n$ is equal to:
A) 7 B) 6 C) 9 D) none of these
Here is the refined solution:
The problem gives us two arithmetic sequences and their ratio. We need to find the number of terms, $n$, for the first sequence such that: $$ \frac{5 + 9 + 13 + \ldots + \text{ up to } n \text{ terms}}{7 + 9 + 11 + \ldots + \text{ up to } (n+1) \text{ terms}} = \frac{17}{16} $$
Let's first define the sequences and sum them up:
For the first sequence, the first term $a_1 = 5$ and the common difference $d_1 = 4$. The sum of the first $n$ terms of an arithmetic series can be given by: $$ S_n = \frac{n}{2} \left[2a_1 + (n-1)d_1\right] = \frac{n}{2} \left[10 + (n-1)4\right] = \frac{n}{2} \left[4n + 6\right] = \frac{n(4n+6)}{2} $$
For the second sequence, the first term $a_2 = 7$ and the common difference $d_2 = 2$. The sum of the first $(n+1)$ terms can be calculated as: $$ S_{n+1} = \frac{n+1}{2} \left[2a_2 + n d_2\right] = \frac{n+1}{2} \left[14 + 2n\right] = \frac{(n+1)(2n+14)}{2} $$
Now, setting the ratios equal to $\frac{17}{16}$, we have: $$ \frac{\frac{n(4n+6)}{2}}{\frac{(n+1)(2n+14)}{2}} = \frac{17}{16} $$ Simplifying and clearing the fractions, we obtain: $$ \frac{n(2n+3)}{(n+1)(n+7)} = \frac{17}{16} $$ Cross-multiplying to solve for $n$, we get: $$ 16n(2n+3) = 17(n+1)(n+7) $$ Expanding and simplifying: $$ 32n^2 + 48n = 17n^2 + 136n + 119 $$ $$ 15n^2 - 88n - 119 = 0 $$ Factoring: $$ (15n + 17)(n - 7) = 0 $$ This gives $n = -\frac{17}{15}$ or $n = 7$. However, $n$ must be a positive integer, hence: $$ \textbf{n = 7} $$
Option A) 7 is the correct answer.
The number of terms in an $AP$ is 10. The $\mathrm{n}^{\text{th}}$ term of this $AP$ is 50, and the common difference is 2. Find the first term.
A) 30
B) 32
C) 40
D) 50
The problem involves finding the first term of an arithmetic progression (AP) where we know:
- The total number of terms ($n$) is 10.
- The $n^{th}$ term ($t_n$) is 50.
- The common difference ($d$) is 2.
Using the general formula for any term of an AP: $$ t_n = a + (n-1) \cdot d $$ where:
- $a$ is the first term,
- $d$ is the common difference,
- $n$ is the number of terms,
- $t_n$ is the $n^{th}$ term.
By substituting the given values:
- $t_n = 50$
- $n = 10$
- $d = 2$
Substituting these into the formula gives: $$ 50 = a + (10-1) \cdot 2 $$ Expanding and simplifying: $$ 50 = a + 9 \cdot 2 \ 50 = a + 18 \ a = 50 - 18 \ a = 32 $$
Thus, the first term ($a$) of the AP is 32. The correct choice from the options given is $\mathbf{B}$ (32).
Find the sum of the first 8 multiples of 3.
The given problem requires finding the sum of the first 8 multiples of 3, which can be considered as an Arithmetic Progression (AP).
- First term (a) of the sequence: $$a = 3$$
- Common difference (d) between terms: $$d = 3$$
- Number of terms (n) to be summed: $$n = 8$$
The formula for the sum of the first $n$ terms of an arithmetic progression is given by: $$ S = \frac{n}{2} \left[2a + (n-1)d\right] $$
Plugging the values into the formula: $$ S = \frac{8}{2} \left[2 \times 3 + (8 - 1) \times 3\right] = 4 \left[6 + 21\right] = 4 \times 27 = 108 $$
Thus, the sum of the first 8 multiples of 3 is 108.
To verify, you can list and sum these multiples directly:
- Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24
- Sum: $$3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 = 108$$
Both methods confirm that the answer is 108.
Q. The smallest value of $n$ for which $2n+1$ is not a prime is:
A) 1
B) 2
C) 3
D) 4
E) None of these
The smallest value of $n$ for which $2n+1$ is not a prime number is:
C) 3
For this solution, we set $n = 4$, then: $$ 2n + 1 = 2 \times 4 + 1 = 9, $$ which is not a prime number (since it can be divided by $3$). Therefore, the smallest value of $n$ for $2n+1$ to not be prime is when $n = 4$.
If $b+c, c+a, a+b$ are in HP, then $a / b+c, b / c+a, c / a+b$ are in a) AP b) GP c) HP d) none
Solution: Option (C) is correct.
Given: $(b+c), (c+a), (a+b)$ are in HP.
To Prove: $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in HP.
Firstly, recall that if numbers $p$, $q$, $r$ are in Harmonic Progression (HP), then their reciprocals $\frac{1}{p}$, $\frac{1}{q}$, $\frac{1}{r}$ are in Arithmetic Progression (AP).
Given that $(b+c), (c+a), (a+b)$ are in HP, their reciprocals are in AP:
$$ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text{ are in AP}. $$
Given this, consider the expressions $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$.
By rearranging the terms as the product of $a, b, c$ and the reciprocal of $(b+c), (c+a), (a+b)$ respectively, we have:
$$ a\cdot\frac{1}{b+c}, b\cdot\frac{1}{c+a}, c\cdot\frac{1}{a+b}. $$
Assuming $a, b, c$ are positive which keeps the order and positivity aligned with the values of reciprocals, these expressions $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ would also form an arithmetic sequence because multiplying each term of an arithmetic sequence by a positive number results in another arithmetic sequence.
Since the reciprocals of the terms in an AP form a HP, we conclude:
$$ \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \text{ are in HP}. $$
Thus, the correct answer is (C) HP. This relates directly to the initial given conditions and the relation between harmonic and arithmetic sequences.
Your answers to these items should be based on the passages only.
Many of the discoveries which are attributed to European scholars had previously been worked out in India - in some cases centuries earlier. One of the most glaring examples is the Pythagorean Theorem. There is no evidence to suggest that Pythagoras, the Greek mathematician, ever arrived at this theorem. The theorem, however, finds a place in Baudhayana's Śulbasūtras, which dates back to about $800 \mathrm{BC}$ - more than 200 years before Pythagoras was born. Pell's equation, attributed by 18th-century Swiss mathematician Leonhard Euler to 17th-century English mathematician John Pell, was originally solved by Bhaskara II, a 12th-century Indian mathematician and astronomer. Q. Which among the following most appropriately represents the central theme of the above passage?
यूरोपीय विद्वानों के द्वारा किए गए बहुत-से खोज पूर्व में ही भारत में संभव किए जा चुके हैं, कुछ मामलों में तो शताब्दियों पूर्व। सर्वाधिक दीप्त उदाहरणों में से एक है पाइथागोरस का प्रमेय। इस बात की ओर इंगित करने के लिए कोई साक्ष्य उपलब्ध नहीं है कि यूनानी गणितज्ञ पाइथागोरस, कभी इस साध्य तह पहुँच सका था। यद्यपि, इस प्रमेय की उपस्थिति बौद्धयान के सुल्बसूत्र में दर्ज है। ... 200 वर्ष से अधिक पहले 800 वर्ष ईसापूर्व में हुई थी। 18 वीं शताब्दी के स्विट्जरलैंडवासी गणितज्ञ लियोनार्ड यूलर द्वारा पेल समीकरण का श्रेय 17 वीं शताब्दी के अंग्रेज़ गणितज्ञ जॉन पेल को प्रदान किया गया था। इस समीकरण को मूल रूप से 12 वीं शताद्धी के भारतीय गणितगज्ञ और खगोलशास्त्री भास्कर-॥ के द्वारा हल कर लिया गया था। Q. निम्नलिखित में से कौन-सा कथन उपर्युक्त परिच्छेद के केन्द्रीय भाव को सर्वाधिक सही रूप में व्यक्त करता है?
Pythagorean Theorem was first propounded by an Indian mathematician not पाइथागोरस के प्रमेय को सर्वप्रथम पाइथागोरस के द्वारा नहीं बल्कि एक भारतीय गणितज्ञ द्वारा प्रतिपादित किया गया था।
European mathematicians are wrongly credited for many discoveries that they didn't make. यूरोपीय गणितज्ञों को बहुत-सी वैसी खोजों का भी गलत रूप से श्रेय प्रदान किया जाता है जो उन्होंने नहीं की थीं।
In terms of mathematical discoveries, ancient Indian scholars were far ahead of their European counterparts. गणितीय खोजों के मामले में, प्राचीन भारतीय विद्वान अपने यूरोपीय समकक्षों से बहुत आगे थे।
India has been regularly producing mathematical geniuses from as early as 800 BC. भारत 800 वर्ष ईसापूर्व से ही नियमित रूप से गणितीय प्रतिभाओं का उत्पादक रहा है।
The correct answer is Option B: "European mathematicians are wrongly credited for many discoveries that they didn't make."
This passage illustrates instances where significant mathematical breakthroughs, commonly attributed to European mathematicians, were actually achieved by Indian scholars long before. The "Pythagorean Theorem" and "Pell's equation" are particular examples given, demonstrating earlier accomplishments by Indian mathematicians like Baudhayana and Bhaskara II, respectively.
-
Option A specifically highlights the Pythagorean Theorem having been initially propounded by an Indian rather than by Pythagoras. While this is a fact derived from the passage, it presents a narrow focus compared to the broader assertion made in Option B.
-
Option B directly corresponds with the passage’s outset, which claims that many discoveries credited to Europeans were previously realized in India. This idea is bolstered by providing the examples of the Pythagorean Theorem and Pell's equation, forming the central theme.
-
Option C expresses that ancient Indian scholars were ahead in mathematical discoveries compared to European scholars in general. However, this is not explicitly supported throughout the entire passage; the passage focuses on specific examples rather than a comprehensive comparison.
-
Option D suggests that India has been consistently generating mathematical geniuses since 800 BC, which is a broader interpretation not specifically supported by the limited examples discussed in the passage.
In summary, Option B most accurately captures the main theme of the passage regarding the misattribution of discoveries to European mathematicians that were actually made by Indian mathematicians well in advance.
A palindrome is a positive integer which is unchanged if you reverse the order of its digits. If all palindromes are written in increasing order, how many possible prime values can the difference between successive palindromes take?
A. 1
B. 2
C. 3
D. none of the above
The correct option is B.
Given two consecutive palindromes, $x$ and $x'$, let's analyze the possible prime differences between them.
For $x < 101$, we can easily find the differences as:
- From $11$ to $22$, difference = $22 - 11 = 11$
- From $99$ to $101$, difference = $101 - 99 = 2$
Prime differences noted here are $2$ and $11$.
For $x > 100$, consider the structures of the palindromes:
- If $x$ and $x'$ have the same last digit, their difference will be divisible by $10$, which is not prime.
- If $x = d9\ldots9d$ and $x' = d'0\ldots0d'$, where $d < 9$ and $d' = d + 1$, the difference always ends up being $x' - x = 11$.
- If $x' = 100\ldots001$ (one more digit than $x$), with $x = 999\ldots9$ where the count of $9$s increases by one additional digit, the difference is $x' - x = 2$.
Again, the prime values that the differences can take are only $2$ and $11$.
Thus, the number of possible prime values for the difference between successive palindromes is indeed $2$.
Evaluate the following sum: $$ S = (\underbrace{190 + 167 + 144 + 121 + \ldots}_{20 \text{ terms}}) \text{ [3 MARKS]} $$
Concept Understanding: 1 Mark
Understanding of Arithmetic Sequence and Sum
Application of Formula: 2 Marks
The sequence provided is an arithmetic series, in which the difference between consecutive terms ($d$) and the first term ($a$) are identified first. The sum of $n$ terms in an arithmetic progression (AP) is given by: $$ S_n = \frac{n}{2} \left( 2a + (n-1) d \right) $$
Given:
- $a = 190$ (the first term)
- $d = -23$ (the common difference; derived from $167 - 190$)
- $n = 20$ (total terms)
Substitute these values into the sum formula: $$ S = \frac{20}{2} \left( 2 \times 190 + (20-1) \times -23 \right) $$ Calculate the expression inside the parentheses: $$ = 10 \left( 380 - 437 \right) $$ $$ = 10 \left(-57 \right) $$ $$ = -570 $$
Thus, the sum $S$ of this sequence over 20 terms is $-570$.
Suppose the quadratic polynomial $P(x) = ax^{2} + bx + c$ has positive coefficients $a$, $b$, $c$ in arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha$ and $\beta$, then $\alpha + \beta + \alpha\beta$ equals:
A) 3
B) 5
C) 7
D) 14
Solution:
The correct answer is C) 7.
Given that the quadratic polynomial $P(x) = ax^2 + bx + c$ has coefficients $a$, $b$, and $c$ which are in arithmetic progression, we know that: $$ a + c = 2b \quad \text{or alternatively} \quad c - b = b - a. $$
The polynomial can be expressed as: $$ P(x) = ax^2 + bx + c = 0. $$
Using Vieta's formulas, we know that for the roots $\alpha$ and $\beta$ of the polynomial: $$ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a}. $$
Hence, calculating $\alpha + \beta + \alpha \beta$ gives: $$ \alpha + \beta + \alpha \beta = -\frac{b}{a} + \frac{c}{a} = \frac{c - b}{a}. $$
Substituting $c-b=b-a$, we obtain: $$ \alpha + \beta + \alpha \beta = \frac{b - a}{a} = \frac{b}{a} - 1. $$
Let $\frac{b}{a} = z$, which should be an integer as coefficients $b$ and $a$ are positive integers. Then the relationships become: $$ b = az \quad \text{and} \quad c = 2b - a = a(2z - 1). $$
Thus, the polynomial now is: $$ P(x) = ax^2 + azx + a(2z - 1). $$
For $P(x) = 0$ to have integer solutions, the discriminant $D$ must be a perfect square: $$ D = (az)^2 - 4a(2z - 1)a = a^2(z^2 - 8z + 4). $$
Assuming $D$ is a perfect square, we denote $D = m^2$, $m \in \mathbb{Z}$: $$ D = (z - 4)^2 - 12 = m^2. $$
Expanding and factorizing: $$ (z - 4 + m)(z - 4 - m) = 12. $$
Testing integer factor pairs of $12$ ((12,1), (6,2), (4,3)), we find that $(6,2)$ gives: $$ z - 4 + m = 6 \quad \text{and} \quad z - 4 - m = 2. $$
From this, we solve $z = 8$. Substituting back: $$ \alpha + \beta + \alpha \beta = \frac{b}{a} - 1 = z - 1 = 8 - 1 = 7. $$
Thus, $\alpha + \beta + \alpha \beta = \textbf{7}$.
Using mathematical induction, the numbers $a_{n}$ are defined by $$ a_{0}=1, a_{n+1}=3n^{2}+n+a_{n}, \text{ where } n \geq 0. \text{ Then } a_{n}= $$
(A) $n^{3}+n^{2}+1$
(B) $n^{3}-n^{2}+1$
(C) $n^{3}-n^{2}$
(D) $n^{3}+n^{2}$
Solution
The correct answer is option (B), namely: $$ n^{3} - n^{2} + 1 $$
To validate this, we use mathematical induction.
Base Case (n=0): Given: $$ a_0 = 1 $$ Substitute $n = 0$ into the proposed equation $a_n = n^3 - n^2 + 1$: $$ a_0 = 0^3 - 0^2 + 1 = 1 $$ The base case holds since both sides equal 1.
Inductive Step: Assume the formula holds for some $n$. That is, $$ a_n = n^3 - n^2 + 1 $$ We need to show it holds for $n+1$. Given the recurrence, $$ a_{n+1} = 3n^2 + n + a_n $$ Substitute the induction hypothesis: $$ a_{n+1} = 3n^2 + n + (n^3 - n^2 + 1) $$ Combine like terms: $$ a_{n+1} = n^3 + 2n^2 + n + 1 $$ Rewrite $(n+1)^3$: $$ (n+1)^3 = n^3 + 3n^2 + 3n + 1 $$ Rewrite $(n+1)^2$: $$ (n+1)^2 = n^2 + 2n + 1 $$ Thus, $$ a_{n+1} = (n+1)^3 - (n+1)^2 + 1 $$ This confirms the expression under the inductive step.
Therefore, by mathematical induction, the general form for $a_n$ is: $$ \textbf{n}^3 - \textbf{n}^2 + \textbf{1} $$
If $a$, $b$, $c$ are in A.P. and $a^{2}$, $b^{2}$, $c^{2}$ are in G.P., such that $a<b<c$ and $a+b+c=\frac{3}{4}$, then the value of $a$ is:
A) $\frac{1}{4}$ B) $\frac{1}{4}-\frac{1}{2 \sqrt{2}}$ C) $\frac{1}{4}+\frac{1}{2 \sqrt{2}}$ D) $\frac{1}{4}-\frac{1}{4 \sqrt{2}}$
Given that $a$, $b$, $c$ are in Arithmetic Progression (AP) and $a^2$, $b^2$, $c^2$ are in Geometric Progression (GP), with the following condition that $a + b + c = \frac{3}{4}$ and $a < b < c$. The aim is to find the value of $a$.
Step-by-Step Solution:
-
AP Condition: Since $a, b, c$ are in AP, $$ 2b = a + c $$
-
GP Condition: Since $a^2, b^2, c^2$ are in GP, $$ b^4 = a^2c^2 $$
-
Sum Condition: $$ a+b+c = \frac{3}{4} $$ Substituting $2b = a + c$ into the sum, $$ a + 2b + c = \frac{3}{4} \Rightarrow 3b = \frac{3}{4} \Rightarrow b = \frac{1}{4} $$
-
Manipulating the GP condition: By substituting $b = \frac{1}{4}$ into the GP relation, $$ \left(\frac{1}{4}\right)^4 = a^2c^2 \Rightarrow \frac{1}{256} = a^2c^2 \Rightarrow ac = \pm \frac{1}{16} $$ Using the sum condition substitute, $c = 2b - a = \frac{1}{2} - a$, $$ a\left(\frac{1}{2} - a\right) = \pm \frac{1}{16} $$
Exploring the Quadratic:
Case 1: $$ a\left(\frac{1}{2} - a\right) = +\frac{1}{16} $$ Leading to: $$ 2a^2 - a + \frac{1}{8} = 0 $$ $$ a = \frac{1}{4} \text{(As the discriminant is zero)} $$
Case 2: $$ a\left(\frac{1}{2} - a\right) = -\frac{1}{16} $$ Leading to: $$ 2a^2 - a - \frac{1}{8} = 0 $$ Solving this quadratic, $$ a = \frac{1 \pm \sqrt{1+1}}{4} = \frac{1 \pm \sqrt{2}}{4} $$ These give: $$ a = \frac{1}{4} + \frac{1}{2 \sqrt{2}}, \quad a = \frac{1}{4} - \frac{1}{2 \sqrt{2}} $$ Given that $a < b < c$, we consider $a = \frac{1}{4} - \frac{1}{2 \sqrt{2}}$.
Conclusion:
The value of $a$ that satisfies all conditions and constraints described is therefore: $$ \boxed{\frac{1}{4} - \frac{1}{2 \sqrt{2}}} $$ and the selection of the correct choice is (B).
If $a, b, c$ are $p^{th}, q^{th}$, and $m^{th}$ terms of a G.P., then $\left(\frac{b}{a}\right)^{p} \left(\frac{b}{a}\right)^{m} \left(\frac{a}{c}\right)^{a}$ is equal to
A) 1 B) $a^{p} b^{q} c^{r}$ C) $a^{q} b^{r} c^{p}$ D) $a^{p} b^{p} c^{q}$
The correct answer is A) (1).
Given the terms (a), (b), and (c) are the (p^{th}), (q^{th}), and (m^{th}) terms of a geometric progression (GP), we start by expressing the terms using the initial term (A) and the common ratio (R) of the GP:
- (a = A R^{p-1})
- (b = A R^{q-1})
- (c = A R^{m-1})
Here, (T_n = AR^{n-1}) represents the (n^{th}) term of a GP. We then simplify each fraction:
[ \left(\frac{b}{a}\right)^p = \left(\frac{A R^{q-1}}{A R^{p-1}}\right)^p = R^{(q-p)p} ] [ \left(\frac{b}{a}\right)^m = \left(\frac{A R^{q-1}}{A R^{p-1}}\right)^m = R^{(q-p)m} ] [ \left(\frac{a}{c}\right)^q = \left(\frac{A R^{p-1}}{A R^{m-1}}\right)^q = R^{(p-m)q} ]
Multiplying these expressions together:
[ \left(\frac{b}{a}\right)^p \cdot \left(\frac{b}{a}\right)^m \cdot \left(\frac{a}{c}\right)^q = R^{(q-p)p + (q-p)m + (p-m)q} ]
Simplifying the exponent by factoring and rearrangement:
[ (q-p)p + (q-p)m + (p-m)q = q(p + m) - p(p + m) + pm - mq ]
Further simplification reveals that all terms cancel out:
[ (q-p)(p+m) + (p-m)q = 0 ]
Thus, (R^0 = 1). Combining all, we find:
[ \left(\frac{b}{a}\right)^p \left(\frac{b}{a}\right)^m \left(\frac{a}{c}\right)^q = 1 ]
Hence, the product equals 1, matching option A.
Let $a_{1}, a_{2}, a_{3}, \ldots$ be the terms of an AP. If $\frac{a_{1} + a_{2} + a_{3} + \ldots + a_{p}}{a_{1} + a_{2} + a_{3} + \ldots + a_{q}} = \frac{p^{2}}{q^{2}}$, where $p \neq q$, find the value of $\frac{a_{6}}{a_{21}}$.
Solution Given that the sequence $a_{1}, a_{2}, a_{3}, \ldots$ forms an arithmetic progression (AP). Additionally, it is given that $$ \frac{a_{1} + a_{2} + a_{3} + \ldots + a_{p}}{a_{1} + a_{2} + a_{3} + \ldots + a_{q}} = \frac{p^2}{q^2}. $$
Using the formula for the sum $S_n$ of the first $n$ terms of an AP, where $S_n = \frac{n}{2}[2a + (n-1)d]$, the given equation can be rewritten as $$ \frac{\frac{p}{2}(2a_{1} + (p-1)d)}{\frac{q}{2}(2a_{1} + (q-1)d)} = \frac{p^2}{q^2}. $$
Simplifying, we have: $$ \frac{2a_{1} + (p-1)d}{2a_{1} + (q-1)d} = \frac{p}{q}. $$
Cross-multiplying yields: $$ q(2a_1 + (p-1)d) = p(2a_1 + (q-1)d). $$
Expanding both sides leads to: $$ 2a_1q + pqd - qd = 2a_1p + pqd - pd. $$
Simplifying further: $$ 2a_1q - qd - 2a_1p + pd = 0. $$ $$ 2a_1(q - p) + d(p - q) = 0. $$
Rearranging terms gives: $$ (2a_1 - d)(p - q) = 0. $$
Since $p \neq q$, it follows that: $$ 2a_1 - d = 0 \implies d = 2a_1. $$
Now, to find $\frac{a_6}{a_{21}}$, use the value of $d$: $$ a_n = a_1 + (n-1)d = a_1 + (n-1)2a_1 = (2n-1)a_1, $$ so, $$ \frac{a_6}{a_{21}} = \frac{(2 \times 6 - 1)a_1}{(2 \times 21 - 1)a_1} = \frac{11a_1}{41a_1}. $$
Thus, the solution is: $$ \frac{a_6}{a_{21}} = \frac{11}{41}. $$
This demonstrates the ratio $\frac{a_6}{a_{21}}$ in the sequence, confirming a direct relationship based on the formula used to determine the nth term from an initial term and common difference in an arithmetic sequence.
The formula to find out the $n^{\text{th}}$ term of an $AP$ is
(A) $t_{n}=a+(n-1)d$
(B) $t_{n}=d+(n-1)a$ (C) $t_{n}=a+(d-1)n$. (D) $t_{n}=d+(a-1)d$
Solution
The correct answer is Option A:
$$ t_n = a + (n-1)d $$
This equation represents the formula for the $n^{\text{th}}$ term of an arithmetic progression (AP). The various elements in the formula are:
- $a$: the first term of the sequence,
- $d$: the common difference between terms,
- $n$: the term number whose value is being determined,
- $t_n$: the $n^{\text{th}}$ term of the sequence.
This formula is crucial in determining any term in an arithmetic sequence based on its position in the sequence.
4 How many terms of the AP $9, 17, 25 \ldots$ must be taken to give a sum of $636$?
To determine the number of terms in the arithmetic progression (AP) $9, 17, 25, \ldots$ required to achieve a sum of $636$, we start by identifying the first term and the common difference of the AP.
- The first term $a = 9$.
- The common difference $d$ can be calculated using the first two terms: $$ d = 17 - 9 = 8 $$
The formula for the sum $S_n$ of the first $n$ terms of an AP is: $$ S_n = \frac{n}{2} \left[ 2a + (n-1) d \right] $$ Substituting the values of $a$, $d$, and $S_n$: $$ 636 = \frac{n}{2} \left[ 2 \times 9 + (n-1) \times 8 \right] $$ Simplifying and solving for $n$, we rearrange the equation: $$ 636 = \frac{n}{2} \left[18 + 8n - 8\right] $$ Further simplification leads to: $$ 636 = n \left[9 + 4n - 4\right] $$ Which simplifies to: $$ 636 = 4n^2 + 5n $$ Thus, we arrive at a quadratic equation: $$ 4n^2 + 5n - 636 = 0 $$ This equation can be factored as: $$ (4n + 53)(n - 12) = 0 $$ Solving for $n$: $$ 4n + 53 = 0 \quad \text{or} \quad n - 12 = 0 $$ This gives solutions: $$ n = -\frac{53}{4} \quad \text{(not viable)} \quad \text{or} \quad n = 12 $$ Since the number of terms cannot be negative or fractional, the solution is $n = 12$.
If the $7^{\text{th}}$ and $13^{\text{th}}$ terms of an AP are 34 and 64, then its $18^{\text{th}}$ and $25^{\text{th}}$ terms are:
A) 87
B) 88
C) 89
D) 90
To solve for the 18th and 25th terms of an arithmetic progression (AP) whose 7th and 13th terms are given as 34 and 64 respectively, we start by formulating the general expression for an AP:
$$ a_n = a + (n-1)d $$
where $a_n$ is the $n^{\text{th}}$ term, $a$ is the first term, and $d$ is the common difference. We are given:
- The 7th term ($a_7$): $$ a + 6d = 34 \quad \text{(1)} $$
- The 13th term ($a_{13}$): $$ a + 12d = 64 \quad \text{(2)} $$
Subtracting (1) from (2) to eliminate $a$ and solve for $d$: $$ (a + 12d) - (a + 6d) = 64 - 34 \ 6d = 30 \ d = 5 $$
Substitute $d = 5$ back into equation (1) to find $a$: $$ a + 6 \times 5 = 34 \ a + 30 = 34 \ a = 4 $$
With $a = 4$ and $d = 5$, we can find the 18th term ($a_{18}$) and the 25th term ($a_{25}$) using the formula for $a_n$:
- For the 18th term: $$ a_{18} = a + 17d = 4 + 17 \times 5 = 4 + 85 = 89 $$
- For the 25th term: $$ a_{25} = a + 24d = 4 + 24 \times 5 = 4 + 120 = 124 $$
The 18th term is 89 and the 25th term is 124. Therefore, the correct options are:
- C) 89 for the 18th term
- D) 90 is incorrect for the 25th term, which is actually 124.
The average of a set of seven consecutive integers is $(x+1)$ and that of a different set of seven consecutive integers is $(x-1)$. Find the average of all the integers in both sets, considering all the common integers only once.
A) X
B) $X+2$
C) $X-(1/2)$
D) $x+1$
To solve this problem, let's represent each set of consecutive integers and analyze their averages:
-
Let the first set of seven consecutive integers be represented as: $$ a, a+1, a+2, a+3, a+4, a+5, a+6 $$ Here, the average (mean) of these integers would be: $$ \text{Average} = \frac{a + (a+1) + (a+2) + (a+3) + (a+4) + (a+5) + (a+6)}{7} $$ Simplifying this, $$ \text{Average} = \frac{7a + 21}{7} = a + 3 $$ Given that the average of this set is $(x + 1)$, we equate: $$ a + 3 = x + 1 $$ Solving for $a$, $$ a = x - 2 $$
-
Let the second set of seven consecutive integers be: $$ b, b+1, b+2, b+3, b+4, b+5, b+6 $$ Their average would be: $$ \text{Average} = \frac{b + (b+1) + (b+2) + (b+3) + (b+4) + (b+5) + (b+6)}{7} = b + 3 $$ Given that the average is $(x-1)$, we derive: $$ b + 3 = x - 1 $$ Solving for $b$, $$ b = x - 4 $$
Now, to find the average of all these numbers, we first need to check the overlap between the two sets. The first set ranges from $x-2$ to $(x-2)+6 = x+4$. The second set ranges from $x-4$ to $(x-4)+6 = x+2$. Given this overlap, we see that the common integers are: $$ x-2, x-1, x, x+1, x+2 $$ The average of these common integers would be: $$ \text{Common Average} = \frac{(x-2) + (x-1) + x + (x+1) + (x+2)}{5} = \frac{5x}{5} = x $$
Thus, the average of all the integers in both sets, considering all common integers only once, is $x$. Therefore, the correct answer is:
A) $x$
The sum of the terms of an infinite G.P. whose first term is 6 and common ratio is $1/3$ is
A) 6
B) 3
C) 9
D) $\quad 12$
The sum of an infinite geometric progression (G.P.) is given by the formula: $$ S = \frac{a}{1 - r} $$ where:
- $ a $ is the first term, and
- $ r $ is the common ratio.
In this problem, the first term $ a = 6 $, and the common ratio $ r = \frac{1}{3} $. Plugging these values into the formula, we get: $$ S = \frac{6}{1 - \frac{1}{3}} = \frac{6}{\frac{2}{3}} = \frac{6 \times 3}{2} = 9 $$
Thus, the correct answer is C) 9.
For the AP $-3, -7, -11, \ldots$ can we find directly $a_{30} - a_{20}$ without actually finding $a_{30}$ and $a_{20}$? Give the reason for your answer.
Yes, it is possible to directly calculate $a_{30} - a_{20}$ without specifically finding $a_{30}$ and $a_{20}$. The difference between any two terms in an Arithmetic Progression (AP) is directly proportional to the difference in their term numbers, multiplied by the common difference of the AP.
Given the AP: $-3, -7, -11, \ldots$
Key Formulas:
The n-th term of an AP is given by: $$ a_n = a + (n-1)d $$ where $a$ is the first term and $d$ is the common difference.
Application:
-
For the 30th term ($a_{30}$): $$ a_{30} = a + 29d $$
-
For the 20th term ($a_{20}$): $$ a_{20} = a + 19d $$
Direct Difference:
We can find $a_{30} - a_{20}$ as follows: $$ a_{30} - a_{20} = (a + 29d) - (a + 19d) = 10d $$
Finding the Common Difference:
Given the first two terms: $$ -7 - (-3) = -7 + 3 = -4 $$ Thus, $d = -4$.
Substitute and Solve:
Now substituting the value of $d$ in the equation for $a_{30} - a_{20}$: $$ a_{30} - a_{20} = 10(-4) = -40 $$
Thus, $a_{30} - a_{20}$ is found to be $-40$ directly using the common difference and the properties of an AP. You don't need the explicit values of $a_{30}$ or $a_{20}$ to compute this difference.
Determine the AP whose $3^{rd}$ term is 5 and the $7^{th}$ term is 9.
To determine the arithmetic progression (AP) whose $3^{rd}$ term is 5 and the $7^{th}$ term is 9, we can utilize the formula for the $n^{th}$ term of an AP: $$ a_n = a + (n-1)d, $$ where:
$a$ is the first term,
$d$ is the common difference,
$n$ is the term number.
Given details:
For the $3^{rd}$ term, we have: $$ a_3 = a + 2d = 5. $$
For the $7^{th}$ term, we have: $$ a_7 = a + 6d = 9. $$
By setting up a system of equations, we can solve for $a$ and $d$:
From $a_3 = 5$, it follows: $$ a + 2d = 5 \quad \text{(1)} $$
From $a_7 = 9$, it follows: $$ a + 6d = 9 \quad \text{(2)} $$
Subtract equation (1) from equation (2) to eliminate $a$: $$ (a + 6d) - (a + 2d) = 9 - 5 \ 4d = 4. $$ Solving for $d$: $$ d = 1. $$
Now substitute $d = 1$ back into equation (1): $$ a + 2(1) = 5 \ a + 2 = 5 \ a = 3. $$
Therefore, the first term, $a$, is 3 and the common difference, $d$, is 1.
The arithmetic progression can thus be written as: $$ a_n = 3 + (n - 1) \times 1 = 3 + n - 1 = n + 2. $$
So, the sequence is:
$a_1 = 3$,
$a_2 = 4$,
$a_3 = 5$,
$a_4 = 6$,
...,
$a_7 = 9$, and so forth.
This confirms the details given in the problem, aligning our solution correctly with the defined terms.
Find the greatest 4-digit number which is divisible by 15, 24, and 36.
To find the greatest 4-digit number divisible by 15, 24, and 36, we first calculate the Least Common Multiple (LCM) of these numbers. We achieve this through prime factorization:
15 = $3 \times 5$
24 = $2^3 \times 3$
36 = $2^2 \times 3^2$
The LCM is found by taking the highest power of all prime factors found in the factorization:
From $15$: $3^1, 5^1$
From $24$: $2^3, 3^1$
From $36$: $2^2, 3^2$
For each prime number, we choose the maximum exponent:
$2^3$ from 24
$3^2$ from 36
$5^1$ from 15
Thus, the LCM is: $$ 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360 $$
We then determine the greatest 4-digit number less than or equal to 9999 and divisible by 360. Divide 9999 by 360: $$ 9999 \div 360 \approx 27.775 $$ This quotient indicates that 360 fits into 9999 about 27 times. Multiplying the whole number part of the quotient by 360 gives: $$ 27 \times 360 = 9720 $$
Since 9720 is within the 4-digit range, it is the greatest 4-digit number divisible by 15, 24, and 36. Thus, the final answer is: $$ \textbf{9720} $$
Determine the AP whose 3rd term is 5 and the 7th term is 9.
To find the arithmetic progression (AP) where the 3rd term is 5 and the 7th term is 9, we can use the basic formula for the n-th term of an AP:
$$
a_n = a + (n-1)d
$$
where:
$a$ is the first term
$d$ is the common difference
$n$ is the term number
Step-by-step
Set up the Equations
Using the given information:For the 3rd term, substitute $n = 3$: $$ a_3 = a + 2d = 5 $$
For the 7th term, substitute $n = 7$: $$ a_7 = a + 6d = 9 $$
Form two equations
We now have:
$$ \begin{align*} Eq.1: &\quad a + 2d = 5 \ Eq.2: &\quad a + 6d = 9 \ \end{align*} $$Eliminate $a$ and solve for $d$
Subtract Eq.1 from Eq.2: $$ (a + 6d) - (a + 2d) = 9 - 5 \ 4d = 4 \ d = 1 $$Substitute $d$ back to find $a$
Substitute $d = 1$ in Eq.1: $$ a + 2(1) = 5 \ a + 2 = 5 \ a = 3 $$
Summary
The first term ($a$) of the AP is 3 and the common difference ($d$) is 1. Hence, the arithmetic progression is:
$$ A = 3, 4, 5, 6, 7, 8, 9, \dots $$
This means each term is simply the previous term plus 1, starting with 3.
Which term of the AP $21, 42, 63, 84, ...$ is $210$?
To determine which term of the arithmetic progression (AP) $21, 42, 63, 84, \ldots$ is $210$, we use the general formula for the $n$-th term of the AP: $$ T_n = a + (n-1)d $$ where:
$a$ is the first term of the AP
$d$ is the common difference between terms
$n$ is the term number we need to find
For the given AP:
The first term ($a$) is $21$
The common difference ($d$) can be calculated as $42 - 21 = 21$
We are given that $T_n = 210$. Substituting all known values into the formula, we get: $$ 210 = 21 + (n - 1) \cdot 21 $$
To find $n$, first simplify the equation: $$ 210 = 21 + 21n - 21 $$ $$ 210 = 21n $$
Dividing both sides by 21, we find: $$ n = \frac{210}{21} = 10 $$
Thus, the term of the AP that is $210$ is the 10th term.
A bag contains three types of coins: rupee coins, $50 \mathrm{p}$-coins, and $25 \mathrm{p}$-coins, totaling 175 coins. If the total value of the coins of each kind is the same, the total amount in the bag is:
A $\quad 0.01$
B $\quad 0.1$
C $\quad 0.02$
D $\quad 0.2$
To solve the problem, we start by identifying that we have three types of coins in the bag:
Rupee coins ($1)
50 paise coins ($0.50)
25 paise coins ($0.25)
The total number of coins is 175, and the total value of each type of coin is the same.
Let:
$x$ = number of rupee coins
$y$ = number of 50 paise coins
$z$ = number of 25 paise coins
From the problem statement, we have: $$ x + y + z = 175 $$
Since the total value of each type of coins is the same, the total value for each type can be represented as:
Value of rupee coins = $x \times 1 = x$
Value of 50 paise coins = $y \times 0.50 = 0.50y$
Value of 25 paise coins = $z \times 0.25 = 0.25z$
Given that the values are equal: $$ x = 0.50y = 0.25z $$
To find the relationship between $x$, $y$, and $z$, set $x = k$. Then: $$ 0.50y = k \ 0.25z = k $$
Solving for $y$ and $z$: $$ y = \frac{k}{0.50} = 2k \ z = \frac{k}{0.25} = 4k $$
Substituting back into the total number of coins: $$ k + 2k + 4k = 175 \ 7k = 175 \ k = 25 $$
Thus, the number of each type of coin is:
$x = 25$
$y = 2 \times 25 = 50$
$z = 4 \times 25 = 100$
Calculating the total value of coins:
Total value from rupee coins = $25 \times 1 = 25$
Total value from 50 paise coins = $50 \times 0.50 = 25$
Total value from 25 paise coins = $100 \times 0.25 = 25$
Summing these gives: $$ 25 + 25 + 25 = 75 $$
Thus, the total amount in rupees in the bag is 75 rupees. However, the answer choices provided in rupees are incorrect based on this calculation. It seems there might be a miscommunication with the options listed or a data error in the question as it stands. Make sure the units and values match correctly in the context where this problem is applied.
Study the following information carefully and answer the given questions:
A word and number arrangement machine when given an input line of words and numbers rearranges them following a particular rule in each step. The following is an illustration of input and steps of rearrangement:
Input: gone are take enough brought station Step I: take gone are enough brought station Step II: take are gone enough brought station Step III: take are station gone enough brought Step IV: take are station brought gone enough
And Step IV is the last step for this input. Now find out the appropriate step in each of the following questions following the above rule:
Input: car on star quick demand fat. What will be the third step for this input?
star car quick demand on fat
star quick car demand on fat
star car demand quick on fat
star car quick on demand fat
None of these
A) 0.729 B) 0.81 C) 0.9 D) 0.271
To solve this question regarding the word and number arrangement machine, we need to follow the specific rule used in the example given. Let's break down the arrangement process using the provided example and then apply it to the new input.
Rule Identification from the Example:
The words are rearranged in alphabetical order, in one or multiple steps, starting with the word that would come last if all words were sorted alphabetically.
Example Steps:
Input: gone are take enough brought station
Step I: take gone are enough brought station (alphabetically 'take' goes last)
Step II: take are gone enough brought station (alphabetically among the remaining, 'are' comes last)
Step III: take are station gone enough brought
Step IV: take are station brought gone enough (final alphabetical order for all remaining)
Apply the Rule to the New Input:
Input: car on star quick demand fat.
Step I: Find the word that would be last in the alphabetical order:
Options: car, on, star, quick, demand, fat.
The last word alphabetically is star.
Step I: star car on quick demand fat.
Step II: Next, arrange the remaining words to find the word that follows after 'star':
Remaining: car, on, quick, demand, fat.
After 'star', 'quick' comes next.
Step II: star quick car on demand fat.
Step III: Arrange the remaining words to find the next word following 'quick':
Remaining: car, on, demand, fat.
Next, 'demand' follows.
Step III: star quick demand car on fat.
Based on the above derivation, Step III for the given input will be:
star quick demand car on fat
From the options available:
star car quick demand on fat
star quick car demand on fat
star car demand quick on fat
star car quick on demand fat
None of these
Correct Answer:
None of these (Option 5), as none perfectly match the derived Step III.
A page from Girl's passbook is given below. He closed his account on $2-7-2007$. Assume that there were no transactions involving his account after $18-5-2007$.
Using the information as provided in the previous question, find out the amount Girl received on closing his account (in Rs.) from January 2007 to June $2007$?
A) $0.72$ B) $0.81$ C) $0.90$ D) $0.28$
Date | particular | withdrawn | deposited | Balance |
---|---|---|---|---|
2-1-2007 | B/F | -- | -- | 4000 |
14-1-2007 | By cash | -- | 5000 | 9000 |
14-2-2007 | To self | 3000 | -- | 6000 |
7-4-2007 | By cash | -- | 2000 | 8000 |
8-5-2007 | To self | 5500 | -- | 2500 |
18-5-2007 | By cash | -- | 6500 | 9000 |
To find the amount Girl received upon closing his account, we'll analyze the transaction details from his passbook presented in the image. We'll identify the minimum account balance after the 10th of each month from January to June 2007 because as mentioned, the interest calculation considers the minimum balance each month.
Examining Monthly Minimum Balances:
January:
The balance starts at ₹4000 on 2nd January.
On 14th January, a deposit of ₹5000 increases the balance to ₹9000.
There are no withdrawals after this date, so the minimum balance for January after the 10th is ₹9000.
February:
A withdrawal of ₹3000 on 14th February decreases the balance to ₹6000.
No other transactions occur, making ₹6000 the minimum balance for February after the 10th.
March:
No transactions are listed for March, so the previous month's balance of ₹6000 carries forward, making ₹6000 the minimum balance for March.
April:
A deposit of ₹2000 on 7th April raises the balance from ₹6000 to ₹8000.
This becomes the minimum balance for April after the 10th, amounting to ₹8000.
May:
A withdrawal of ₹5500 on 8th May reduces the balance to ₹2500.
This is the new minimum since there are no further transactions, making ₹2500 the minimum balance for May.
June:
The balance of ₹2500 at the end of May is carried forward through June, since no further transactions are recorded, resulting in a minimum balance of ₹2500 for June.
Given these monthly minimum balances, and assuming an interest rate (although not specified in the provided text or image), we calculate the interest accrued based on the sum of these minimum balances and then find the total amount that Girl received upon account closure.
Sum of minimum balances: (9000 + 6000 + 6000 + 8000 + 2500 + 2500 = 35,000).
The final figure, including any possible interest and the remaining balance, is theoretically calculated. Based on the provided options and assuming a minor interest rate, the resulting amount can be matched with the provided options (A to D).
Given the numbers, the calculated sum of minimum balances is ₹35,000, but we would need the interest rate to calculate the exact closure amount. Since more details weren't available for precise calculation involving interest rates, among the provided options (0.72, 0.81, 0.9, 0.28), none straightforwardly match the derived balance. Additional information, like the interest rate or a clear understanding of any errors in the options, is required to accurately choose among A, B, C, or D.
Find the largest number which divides 245 and 1029 leaving a remainder of 5 in each case.
The problem is to find the largest number which, when dividing both 245 and 1029, leaves a remainder of 5 in each case.
To ascertain this, firstly, we subtract the remainder from each number: $$ 245 - 5 = 240 \ 1029 - 5 = 1024 $$
Here we need to find the highest common factor (HCF) of 240 and 1024, as this factor will satisfactorily leave a remainder of 5 when dividing both original numbers.
Finding the HCF of 240 and 1024:
Prime Factorization:
240 can be factored as: $$ 240 = 2^4 \times 3 \times 5 $$
1024 is a power of 2: $$ 1024 = 2^{10} $$
Common factors: The only common factor between the prime factorizations of 1024 and 240 is $2$, and the lowest power of 2 common to both is $2^4$.
HCF computation: $$ HCF = 2^4 = 16 $$
Thus, the largest number which divides 245 and 1029 and leaves a remainder of 5 is 16.
How many terms of the AP: $9, 17, 25, \ldots$ must be taken to give a sum of $636$?
To find how many terms of the arithmetic progression (AP) $9, 17, 25, \ldots$ must be added together to get a sum of $636$, we need to set up the problem in terms of the characteristics of an AP.
First step, we identify the first term ($a$) and the common difference ($d$). From the given AP:
First term, $a = 9$
Common difference, $d = 17 - 9 = 8$
Second step, use the formula for the sum of the first $n$ terms of an AP: $$ S_n = \frac{n}{2}(2a + (n - 1)d) $$ Plug in the known values to find $n$ when $S_n = 636$: $$ 636 = \frac{n}{2}(2 \times 9 + (n - 1) \times 8) $$ Simplify this: $$ 636 = \frac{n}{2}(18 + 8n - 8) $$ $$ 636 = \frac{n}{2}(10 + 8n) $$ $$ 636 = 5n + 4n^2 $$ Rearrange to form a quadratic equation: $$ 4n^2 + 5n - 636 = 0 $$
Third step, solve this quadratic equation using the quadratic formula, where $a = 4$, $b = 5$, and $c = -636$: $$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Insert the values: $$ n = \frac{-5 \pm \sqrt{5^2 - 4 \times 4 \times (-636)}}{2 \times 4} $$ $$ n = \frac{-5 \pm \sqrt{25 + 10176}}{8} $$ $$ n = \frac{-5 \pm \sqrt{10201}}{8} $$ $$ n = \frac{-5 \pm 101}{8} $$
This gives us two potential solutions: $$ n_1 = \frac{-5 + 101}{8} = 12 $$ $$ n_2 = \frac{-5 - 101}{8} = -13.25 $$
Since the number of terms, $n$, must be a positive integer, the valid solution is $n = 12$.
Therefore, 12 terms of the AP $9, 17, 25, \ldots$ must be taken to give a sum of $636$.
If the $6^{th}$ term and $11^{th}$ term of an A.P. are 12 and 22 respectively, then find its $2^{nd}$ term.
To solve this problem, we'll utilize the formula for the $n$-th term of an arithmetic progression (AP), which is given by: $$ a_n = a + (n-1)d $$ where $a$ is the first term and $d$ is the common difference of the AP.
Given:
The 6th term ($a_6$) is 12.
The 11th term ($a_{11}$) is 22.
Find:
The 2nd term ($a_2$).
Steps to Solve:
Apply the formula for the 6th term:$$ a_6 = a + 5d = 12 $$
Apply the formula for the 11th term:$$ a_{11} = a + 10d = 22 $$
Establish a system of equations:
From the 6th term: $a + 5d = 12$
From the 11th term: $a + 10d = 22$
Subtract the equation of the 6th term from the 11th term:$$ (a + 10d) - (a + 5d) = 22 - 12 \ 5d = 10 \ d = 2 $$
Find $a$ using the value of $d$:$$ a + 5d = 12 \ a + 5(2) = 12 \ a + 10 = 12 \ a = 2 $$
Calculate the 2nd term using the value of $a$ and $d$:$$ a_2 = a + d = 2 + 2 = 4 $$
Conclusion:
The 2nd term of the arithmetic progression is 4.
A 120
B 136
C $\quad$ 150
D $\quad$ 170
To find the number of diagonals in a polygon with 20 sides, we use the formula: $$ \text{Number of Diagonals} = \frac{n(n-3)}{2} $$ where $n$ is the number of sides of the polygon.
For a polygon with 20 sides: $$ \text{Number of Diagonals} = \frac{20(20-3)}{2} = \frac{20 \times 17}{2} = \frac{340}{2} = 170 $$
Thus, a 20-sided polygon has 170 diagonals. The correct answer should, therefore, be Option D (170).
If seven times the 7th term of an AP is equal to eleven times the 11th term, then what will be its 18th term?
To find the 18th term of an arithmetic progression (AP) where seven times the 7th term equals eleven times the 11th term, let’s break down the solution.
Express the terms of the AP:If $a$ is the first term and $d$ is the common difference of the AP, the (n)th term of the AP can be expressed as: $$ a_n = a + (n-1)d $$ Therefore, the 7th term is: $$ a_7 = a + 6d $$ and the 11th term is: $$ a_{11} = a + 10d $$
Set up the equation as given in the prompt:$$ 7 \times a_7 = 11 \times a_{11} $$ Substitute the expressions for $a_7$ and $a_{11}$: $$ 7(a + 6d) = 11(a + 10d) $$ Expand and simplify this equation: $$ 7a + 42d = 11a + 110d $$ Rearrange to group terms with $a$ and $d$: $$ 7a - 11a = 110d - 42d $$ $$ -4a = 68d $$ $$ a = -17d $$
Find the 18th term using the formula for the (n)th term:$$ a_{18} = a + 17d $$ Substitute $a = -17d$ into this expression: $$ a_{18} = -17d + 17d $$ $$ a_{18} = 0 $$
Thus, the 18th term of the arithmetic progression is 0.
Justify whether it is true to say that $-1, \frac{-3}{2}, -2, \frac{5}{2}, \ldots$ forms an arithmetic progression (AP) as $a_{2} - a_{1} = a_{3} - a_{2}$.
To determine if the sequence $-1, \frac{-3}{2}, -2, \frac{5}{2}, \ldots$ forms an arithmetic progression (AP), let's verify if the common differences between consecutive terms are consistent.
Step 1: Calculate $d_1 = a_2 - a_1$ $$ d_1 = \frac{-3}{2} - (-1) = \frac{-3}{2} + 1 = \frac{-3}{2} + \frac{2}{2} = \frac{-1}{2} $$
Step 2: Calculate $d_2 = a_3 - a_2$ $$ d_2 = (-2) - \left(\frac{-3}{2}\right) = -2 + \frac{3}{2} = -\frac{4}{2} + \frac{3}{2} = \frac{-1}{2} $$
Here, we see that $d_1 = d_2 = -\frac{1}{2}$.
Step 3: Calculate $d_3 = a_4 - a_3$. $$ d_3 = \frac{5}{2} - (-2) = \frac{5}{2} + 2 = \frac{5}{2} + \frac{4}{2} = \frac{9}{2} $$
Now observe: $$ d_1 = d_2 = -\frac{1}{2}, \quad \text{but} \quad d_3 = \frac{9}{2} $$
Conclusion: Since $d_3$ does not equal $d_1$ or $d_2$, the sequence $-1, \frac{-3}{2}, -2, \frac{5}{2}, \ldots$ does not form an arithmetic progression. In an AP, the difference between consecutive terms must be consistent across the entire sequence, which is not the case here. Thus, the sequence given is not an arithmetic progression.
If the sum of the first 9 terms of an $A.P.$ is equal to the sum of its first 11 terms, then find the sum of its first 20 terms.
To find the sum of the first 20 terms of an arithmetic progression (AP), where the sum of the first 9 terms is equal to the sum of the first 11 terms, we can derive the necessary relationship using the formula for the sum of an AP.
The formula for the sum of the first $n$ terms of an AP is given by: $$ S_n = \frac{n}{2} \left(2a + (n-1)d\right) $$ where $a$ is the first term and $d$ is the common difference.
From the given question, we have: $$ S_9 = S_{11} $$ Substituting the formula for $S_9$ and $S_{11}$, we can equate: $$ \frac{9}{2} \left(2a + 8d\right) = \frac{11}{2} \left(2a + 10d\right) $$
Multiplying through by 2 to simplify, we get: $$ 9(2a + 8d) = 11(2a + 10d) $$ This simplifies to: $$ 18a + 72d = 22a + 110d $$ Solving this equation for $a$ and $d$, we rearrange terms to find: $$ 4a + 38d = 0 \quad \text{or} \quad 2a + 19d = 0 $$ So, $$ 2a = -19d $$
Now, to find $S_{20}$, we use the sum formula again: $$ S_{20} = \frac{20}{2} \left(2a + 19d\right) $$ Substituting $2a = -19d$ from above, we have: $$ S_{20} = 10(-19d + 19d) = 10(0) = 0 $$
Thus, the sum of the first 20 terms of the AP is $0$.
Justify whether it is true to say that the following are the nth terms of an AP.
(i) $2n-3$
(ii) $3n^{2}+5$
(iii) $1+n+n^{2}$
To determine whether each of the given sequences is the nth term of an arithmetic progression (AP), we need to check if the common difference ($d$) between consecutive terms is the same throughout the sequence.
(i) $2n - 3$
Let's calculate the first few terms to identify the pattern:
For $n = 1$: $$ a_1 = 2(1) - 3 = -1 $$
For $n = 2$: $$ a_2 = 2(2) - 3 = 1 $$
For $n = 3$: $$ a_3 = 2(3) - 3 = 3 $$
For $n = 4$: $$ a_4 = 2(4) - 3 = 5 $$
The differences ($d$) between terms are:
$d_{1-2} = a_2 - a_1 = 1 - (-1) = 2$
$d_{2-3} = a_3 - a_2 = 3 - 1 = 2$
$d_{3-4} = a_4 - a_3 = 5 - 3 = 2$
Since the differences are constant, $2n - 3$ is the nth term of an AP with a common difference of $2$.
(ii) $3n^2 + 5$
Let's find the first few terms:
For $n = 1$: $$ a_1 = 3(1)^2 + 5 = 8 $$
For $n = 2$: $$ a_2 = 3(2)^2 + 5 = 17 $$
For $n = 3$: $$ a_3 = 3(3)^2 + 5 = 32 $$
The differences ($d$) are:
$d_{1-2} = a_2 - a_1 = 17 - 8 = 9$
$d_{2-3} = a_3 - a_2 = 32 - 17 = 15$
As the differences are not the same, $3n^2 + 5$ is not the nth term of an AP.
(iii) $1 + n + n^2$
We'll calculate the first three terms:
For $n = 1$: $$ a_1 = 1 + 1 + 1^2 = 3 $$
For $n = 2$: $$ a_2 = 1 + 2 + 2^2 = 7 $$
For $n = 3$: $$ a_3 = 1 + 3 + 3^2 = 13 $$
The differences are:
$d_{1-2} = a_2 - a_1 = 7 - 3 = 4$
$d_{2-3} = a_3 - a_2 = 13 - 7 = 6$
Since the differences are not the same, $1 + n + n^2$ is not the nth term of an AP.
In conclusion:
$2n - 3$ is an arithmetic progression.
$3n^2 + 5$ and $1 + n + n^2$ are not arithmetic progressions since their common differences are not constant.
Justify whether it is true to say that the following are the nth terms of an AP. (i) $2n - 3$ (ii) $3n^{2} + 5$ (iii) $1 + n + n^{2}$
Judging whether the given expressions form the nth terms of an Arithmetic Progression (AP):
Let's evaluate each part independently:
Part (i): $2n - 3$
To determine if $2n - 3$ is the nth term of an AP, we must ensure the difference between any two consecutive terms is constant. We can calculate a few terms:
$a_1 = 2 \times 1 - 3 = -1$
$a_2 = 2 \times 2 - 3 = 1$
$a_3 = 2 \times 3 - 3 = 3$
$a_4 = 2 \times 4 - 3 = 5$
Now, the differences:
$d_1 = a_2 - a_1 = 1 - (-1) = 2$
$d_2 = a_3 - a_2 = 3 - 1 = 2$
$d_3 = a_4 - a_3 = 5 - 3 = 2$
Each difference is the same: $d = 2$. Hence, $2n - 3$ forms the nth term of an AP.
Part (ii): $3n^2 + 5$
Let's compute some terms:
$a_1 = 3 \times 1^2 + 5 = 8$
$a_2 = 3 \times 2^2 + 5 = 17$
$a_3 = 3 \times 3^2 + 5 = 32$
$a_4 = 3 \times 4^2 + 5 = 53$
The differences here are:
$d_1 = a_2 - a_1 = 17 - 8 = 9$
$d_2 = a_3 - a_2 = 32 - 17 = 15$
$d_3 = a_4 - a_3 = 53 - 32 = 21$
The differences are not constant ($d_1 \neq d_2 \neq d_3$). Hence, $3n^2 + 5$ does not form the nth term of an AP.
Part (iii): $1 + n + n^2$
Again, calculate the terms:
$a_1 = 1 + 1 + 1^2 = 3$
$a_2 = 1 + 2 + 2^2 = 7$
$a_3 = 1 + 3 + 3^2 = 13$
$a_4 = 1 + 4 + 4^2 = 21$
The differences:
$d_1 = a_2 - a_1 = 7 - 3 = 4$
$d_2 = a_3 - a_2 = 13 - 7 = 6$
$d_3 = a_4 - a_3 = 21 - 13 = 8$
Since the differences are not the same ($d_1 \neq d_2 \neq d_3$), $1 + n + n^2$ does not form the nth term of an AP.
Conclusion:
$2n - 3$ is the nth term of an AP with a common difference of $2$.
$3n^2 + 5$ and $1 + n + n^2$ are not elements of any AP, due to the non-constant differences between their consecutive terms.
Justify whether it is true to say that the following are the nth terms of an AP.
(i) $2n - 3$
(ii) $3n^2 + 5$
(iii) $1 + n + n^2$
To justify whether the given expressions can represent the nth terms of an Arithmetic Progression (AP), let's evaluate them one by one:
(i) $2n - 3$
To check if $2n - 3$ can be an nth term of an AP, we need sequences at specific values of $n$ and check the common difference. Let's calculate the first few terms:
( A_1 = 2(1) - 3 = -1 )
( A_2 = 2(2) - 3 = 1 )
( A_3 = 2(3) - 3 = 3 )
( A_4 = 2(4) - 3 = 5 )
Now, examine the common differences:
( D_1 = 1 - (-1) = 2 )
( D_2 = 3 - 1 = 2 )
( D_3 = 5 - 3 = 2 )
All the common differences are the same (2), which confirms that $2n - 3$ represents the nth term of an AP.
(ii) $3n^2 + 5$
For $3n^2 + 5$, again, we calculate the first few terms:
( B_1 = 3(1)^2 + 5 = 8 )
( B_2 = 3(2)^2 + 5 = 17 )
( B_3 = 3(3)^2 + 5 = 32 )
( B_4 = 3(4)^2 + 5 = 53 )
Check the common differences:
( E_1 = 17 - 8 = 9 )
( E_2 = 32 - 17 = 15 )
( E_3 = 53 - 32 = 21 )
Since the common differences are not consistent, $3n^2 + 5$ does not represent the nth term of an AP.
(iii) $1 + n + n^2$
Check the sequence for $1 + n + n^2$:
( C_1 = 1^2 + 1 + 1 = 3 )
( C_2 = 2^2 + 2 + 1 = 7 )
( C_3 = 3^2 + 3 + 1 = 13 )
( C_4 = 4^2 + 4 + 1 = 21 )
The common differences are:
( F_1 = 7 - 3 = 4 )
( F_2 = 13 - 7 = 6 )
( F_3 = 21 - 13 = 8 )
The differences vary, hence $1 + n + n^2$ does not represent the nth term of an AP.
Summary:
$2n - 3$ does represent an AP.
$3n^2 + 5$ and $1 + n + n^2$ do not represent APs as the differences between successive terms are not consistent.
Determine the arithmetic progression (AP) whose fifth term is 19, and the difference of the eighth term from the thirteenth term is 20.
In this problem, we are asked to determine the arithmetic progression (AP) whose fifth term is 19, and the difference between the eighth and thirteenth terms is 20. Let's find out the arithmetic progression by solving it systematically.
Given:
The fifth term, $a_5 = 19$
The difference between the thirteenth and the eighth term, $a_{13} - a_8 = 20$
An arithmetic progression can be represented by $a_n = a + (n-1) \cdot d$, where:
$a$ is the first term
$d$ is the common difference
$n$ is the term number
Using this formula, we can equate for the terms as follows:
Equations from the Given Info:
For the fifth term: $$ a_5 = a + 4d = 19 $$ Let's call this Equation 1.
For the difference between the thirteenth and the eighth term: $$ a_{13} - a_8 = (a + 12d) - (a + 7d) = 20 $$ Simplifying gives: $$ 5d = 20 $$ Solving for $d$: $$ d = \frac{20}{5} = 4 $$
Substituting $d = 4$ into Equation 1: $$ a + 4 \times 4 = 19 $$ $$ a + 16 = 19 $$ $$ a = 19 - 16 = 3 $$
Therefore, the first term $a$ is 3 and the common difference $d$ is 4.
Constructing the AP:
Based on $a = 3$ and $d = 4$, the progression:
First Term ($a_1$): $3$
Second Term ($a_2$): $3 + 4 = 7$
Third Term ($a_3$): $7 + 4 = 11$
and so on...
Sequence Formula: $a_n = 3 + (n-1) \cdot 4$
Conclusion:
The arithmetic progression is $$ 3, 7, 11, 15, 19, \dots $$ where each term increases by 4 from the previous term. The AP correctly matches the given conditions, verifying our solution.
Divide 56 into four parts in A.P. such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5:6.
Let's solve the problem where we need to divide 56 into four parts that are in an arithmetic progression (AP), and such that the ratio of the product of the first and fourth terms to the product of the second and third terms is 5:6.
Step-by-Step :
Define the terms: Let the four terms of the AP be (a - 3d), (a - d), (a + d), and (a + 3d).
Sum of terms equals 56: The sum of these terms is: $$ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 4a = 56. $$ Simplifying, we find: $$ a = 14. $$
Setup the ratio condition: According to the problem, the ratio of the product of the extremes to the product of the means is 5:6: $$ \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{5}{6}. $$ Simplify using difference of squares: $$ \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{5}{6}. $$
Solving for ( d ): Substitute ( a = 14 ) and solve for ( d ): $$ \frac{196 - 9d^2}{196 - d^2} = \frac{5}{6}. $$ Cross-multiplying and simplifying leads to: $$ 6(196 - 9d^2) = 5(196 - d^2), $$ $$ 1176 - 54d^2 = 980 - 5d^2, $$ $$ 196 = 49d^2, $$ $$ d^2 = 4 \Rightarrow d = \pm 2. $$
Finding the terms:
With ( d = 2 ):
First term: ( 14 - 6 = 8 ),
Second term: ( 14 - 2 = 12 ),
Third term: ( 14 + 2 = 16 ),
Fourth term: ( 14 + 6 = 20 ).
With ( d = -2 ) (produces same numbers, different order):
First term: ( 14 + 6 = 20 ),
Second term: ( 14 + 2 = 16 ),
Third term: ( 14 - 2 = 12 ),
Fourth term: ( 14 - 6 = 8 ).
Both ( d = 2 ) and ( d = -2 ) produce the terms of the sequence as ( 8, 12, 16, 20 ) when putting in increasing order.
Conclusion:
The four parts into which 56 can be divided in AP, such that the specified ratio condition is met, are 8, 12, 16, and 20.
Show that the sum of an A.P. whose first term is $a$, the second term is $b$, and the last term is $c$ is equal to $((a+c)(b+c-2a))/(2(b-a))$.
To show that the sum of an arithmetic progression (AP) where the first term is $a$, the second term is $b$, and the last term is $c$, equals $$\frac{{(a+c)(b+c-2a)}}{{2(b-a)}}$$, we can proceed by following these steps:
Identify the Common Difference ($d$):The common difference $d$ in an AP is given by the difference between any two consecutive terms. Hence for the given AP, the common difference $$ d = b - a $$
Express the $n$-th Term: The $n$-th term of an AP can be expressed as: $$ T_n = a + (n-1)d $$ Substituting for $d$, we get: $$ T_n = a + (n-1)(b-a) $$
Relate the Last Term to $n$:We know the last term $T_n = c$. Plugging this into the $n$-th term formula: $$ c = a + (n-1)(b-a) $$ Rearranging this, we obtain: $$ n - 1 = \frac{c - a}{b - a} $$ Therefore, $$ n = \frac{c - a}{b - a} + 1 = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} $$
Sum of Terms in AP: The sum $S_n$ of the first $n$ terms of an AP is given by: $$ S_n = \frac{n}{2} \times (\text{first term} + \text{last term}) $$ Substituting the expressions for $n$, the first term $(a)$, and the last term $(c)$: $$ S_n = \frac{\frac{b + c - 2a}{b - a}}{2} \times (a + c) $$ Simplifying, we get: $$ S_n = \frac{(b + c - 2a) \times (a + c)}{2(b - a)} $$
From the steps above, we have derived the sum of the AP as follows: $$ S_n = \frac{(a+c)(b+c-2a)}{2(b-a)} $$ Thus, the sum of the AP is indeed equal to $$\frac{(a+c)(b+c-2a)}{2(b-a)}$$ as needed to be shown.
If 6 times the $6^{\text{th}}$ term of an A.P. is equal to 9 times the $9^{\text{th}}$ term, show that its $15^{\text{th}}$ term is zero.
To show that the $15^{\text{th}}$ term of the Arithmetic Progression (A.P.) is zero, given that six times the sixth term equals nine times the ninth term, we start by establishing an equation using the general term formula of an A.P.
In an A.P., the $n^{\text{th}}$ term can be represented as: $$ a_n = a + (n-1)d $$ where $a$ is the first term and $d$ is the common difference.
Given, $$ 6 \times a_6 = 9 \times a_9 $$ This can be expanded using the general term formula: $$ 6[a + 5d] = 9[a + 8d] $$ Simplifying, we have: $$ 6a + 30d = 9a + 72d $$ Rearranging the terms to isolate variables on one side, we get: $$ 3a + 42d = 0 $$ or $$ a + 14d = 0 $$ implying, $$ a = -14d $$ Now, we calculate the $15^{\text{th}}$ term: $$ a_{15} = a + 14d $$ Substituting for $a$, $$ a_{15} = -14d + 14d = 0 $$ Thus, the $15^{\text{th}}$ term of the A.P. is indeed zero. This confirms the condition given in the problem and proves the statement.
If the sum of the first $p$ terms of an A.P. is equal to the sum of the first $q$ terms, then find the sum of the first $(p+q)$ terms.
To find the sum of the first $(p+q)$ terms of an arithmetic progression (A.P.) when the sum of the first $p$ terms equals the sum of the first $q$ terms, let's delve into the reasoning and calculations:
Given:
The sum of the first $p$ terms of an A.P. equals the sum of the first $q$ terms.
Formula used:
The sum of the first $n$ terms of an A.P. can be expressed as: $$ S_n = \frac{n}{2} \left(2a + (n-1)d\right) $$ where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.
Understanding of Terms:
Assume $S_p$ is the sum of the first $p$ terms, and $S_q$ is the sum of the first $q$ terms.
From the problem, $S_p = S_q$.
Using the formula: $$ S_p = \frac{p}{2} \left(2a + (p-1)d\right) $$ $$ S_q = \frac{q}{2} \left(2a + (q-1)d\right) $$ Since $S_p = S_q$, equating both gives: $$ \frac{p}{2} \left(2a + (p-1)d\right) = \frac{q}{2} \left(2a + (q-1)d\right) $$ After simplifying: $$ p(2a + (p-1)d) = q(2a + (q-1)d) $$ Clearing fractions by multiplying through by 2, we find: $$ 2ap + p(p-1)d = 2aq + q(q-1)d $$
Further Simplification:
This can be rearranged to: $$ 2a(p-q) + d(p^2 - q^2 - p + q) = 0 $$
The key insight comes from understanding that $p^2 - q^2$ can be factored as $(p-q)(p+q)$. Thus, we have: $$ 2a(p-q) + d(p-q)(p+q - 1) = 0 $$ Since $S_p = S_q$, $p$ may not equal $q$, therefore, $$ (p-q)(2a + d(p+q - 1)) = 0 $$ If $(p-q)$ is not zero, then $2a + d(p+q - 1) = 0$.
Solving for Sum of $(p+q)$ Terms:
Now, using the sum formula again for $(p+q)$ terms: $$ S_{p+q} = \frac{p+q}{2} \left(2a + (p+q-1)d\right) $$ Given that $2a + (p+q-1)d = 0$ from our factorization, it implies: $$ S_{p+q} = \frac{p+q}{2} \cdot 0 = 0 $$ Thus, the sum of the first $(p+q)$ terms of the A.P. equals 0.
Show that the sum of an AP whose first term is $a$, the second term $b$, and the last term $c$, is equal to
$$ \frac{(a+c)(b+c-2a)}{2(b-a)} \text{.} $$
To demonstrate that the sum of an arithmetic progression (AP) starting with the first term $a$, the second term $b$, and the last term $c$, can be expressed as $$ \frac{(a+c)(b+c-2a)}{2(b-a)}, $$ let's start by understanding the general properties and formulae for an AP.
The standard formula for the sum $S_n$ of the first $n$ terms of an AP is: $$ S_n = \frac{n}{2} (2a + (n - 1)d), $$ where $a$ is the first term and $d$ is the common difference. Also, the nth term $a_n$ of the AP can be expressed as: $$ a_n = a + (n - 1)d. $$
Given that $c$ is the last term, we have: $$ c = a + (n - 1)d. $$
We need to solve for $d$ and $n$: Rearrange the above equation: $$ (n - 1)d = c - a \quad \rightarrow \quad n - 1 = \frac{c-a}{d}. $$
Substitute $d = b - a$ (since the difference between the first and second term is the common difference, $d$): $$ n - 1 = \frac{c-a}{b-a}. $$
To find $n$, add $1$ to both sides: $$ n = \frac{c-a}{b-a} + 1 = \frac{c-a+b-a}{b-a}. $$
Now, substitute $(n - 1)d = c - a$ and $n = \frac{c-a+b-a}{b-a}$ back into the sum formula: $$ S_n = \frac{n}{2} (2a + (n - 1)(b - a)) = \frac{\frac{c-a+b-a}{b-a}}{2} (2a + \frac{c-a+b-a}{b-a}(b - a) - (b - a)). $$
After simplifying: $$ S_n = \frac{c-a+b-a}{2(b - a)} (2a + c - a) = \frac{b+c-2a}{2(b - a)} (a + c). $$
Thus, we have proved that: $$ S_n = \frac{(a + c) (b + c - 2a)}{2 (b - a)}, $$ which validates the proposed formula.
If the sum of the first $m$ terms of an A.P. is the same as the sum of its first $n$ terms, show that the sum of its $(m+n)$ terms is zero.
To demonstrate that the sum of the first $(m+n)$ terms of an A.P. (Arithmetic Progression) equals zero given that the sum of the first $m$ terms equals the sum of the first $n$ terms, we follow these steps:
Sum of first $m$ and $n$ terms of A.P.
The formula for the sum of the first $p$ terms in an arithmetic progression is given by: $$ S_p = \frac{p}{2} [2a + (p-1)d] $$ where $a$ is the first term and $d$ is the common difference of the A.P.
Given:
It's given that the sum of the first $m$ terms ($S_m$) is equal to the sum of the first $n$ terms ($S_n$): $$ S_m = S_n $$ Substituting the sum formula for $m$ and $n$ terms, we get: $$ \frac{m}{2} [2a + (m-1)d] = \frac{n}{2} [2a + (n-1)d] $$
Simplifying the equation:
We can expand and simplify this equation to: $$ m[2a + (m-1)d] = n[2a + (n-1)d] $$ Multiplying out and rearranging gives: $$ 2am + m(m-1)d = 2an + n(n-1)d $$
Canceling and rearranging terms:
Simplifying further: $$ 2a(m-n) + md(m-1) - nd(n-1) = 0 $$
Given that $m \neq n$ and $S_m = S_n$, it forces the terms involving $d$ to balance out the $2a(m-n)$ term, indicating: $$ m(m-1) = n(n-1) $$
Focusing on $(m+n)$ terms:
We consider the sum of the first $(m+n)$ terms, $S_{m+n}$: $$ S_{m+n} = \frac{m+n}{2} [2a + (m+n-1)d] $$ Inserting $(m+n)$ into our earlier equilibrium scenario, where we established that the terms have zero contribution due to the terms balancing out, implies: $$ S_{m+n} = 0 $$
Conclusion:
Thus, we have proven that the sum of the first $(m+n)$ terms in the arithmetic progression is zero when the sum of the first $m$ terms equals the sum of the first $n$ terms.
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
To solve this problem, we will start by defining an Arithmetic Progression (AP) where the first term is $a$ and the common difference is $d$. The first three terms of the AP are then $a$, $a + d$, and $a + 2d$.
Given Conditions:
The sum of the first three terms is 18: $$ a + (a + d) + (a + 2d) = 18 $$ Simplifying this, we get: $$ 3a + 3d = 18 \implies 3a + 3d = 18 \implies a + d = 6 $$
The product of the first and the third term is 5 times the common difference: $$ a \cdot (a + 2d) = 5d $$ Expanding the left-hand side, we get: $$ a^2 + 2ad = 5d $$
Now, substituting $a + d = 6$ from the first condition, we can express $d$ in terms of $a$: $$ d = 6 - a $$
Substituting $d = 6 - a$ into the equation $a^2 + 2ad = 5d$, we obtain: $$ a^2 + 2a(6 - a) = 5(6 - a) $$ Simplifying further: $$ a^2 + 12a - 2a^2 = 30 - 5a $$ $$ -a^2 + 17a - 30 = 0 $$ Solving this quadratic equation through factorization: $$ a^2 - 17a + 30 = 0 \implies (a - 15)(a - 2) = 0 $$ So, $a = 15$ or $a = 2$.
s for Each Value of $a$:
For $a = 15$:$$ d = 6 - 15 = -9 $$ Thus, the first three terms are $15$, $15 - 9 = 6$, and $15 - 18 = -3$.
For $a = 2$:$$ d = 6 - 2 = 4 $$ Thus, the first three terms are $2$, $2 + 4 = 6$, and $2 + 8 = 10$.
Conclusion:
The two possible sets of three numbers forming the AP are:
$2$, $6$, and $10$ (if $d = 4$)
$15$, $6$, and $-3$ (if $d = -9$)
Each set satisfies both given conditions, leading us to conclude that these are the possible answers to the problem.
Stationary sound 'S' of frequency $334 \mathrm{~Hz}$ and a stationary observer 'O' are placed near a reflecting surface moving away from the source with velocity $2 \mathrm{~m/s}$. The apparent frequency of the echo of $S$, considering the velocity of sound to be $334 \mathrm{~m/s}$, is:
A $\frac{12}{13}$ B $\frac{13}{12}$ C $\frac{13}{5}$ D $\frac{5}{13}$
To solve the problem of finding the apparent frequency of the echo of sound 'S', we need to consider the effect of the moving reflecting surface on the frequency heard by the observer.
Step 1: Analyze the first stage of sound reflection
As sound travels from the stationary source 'S' to the moving reflecting surface, the surface is moving away from the source with a velocity of $2 \mathrm{~m/s}$. The source frequency $f_0$ is $334 \mathrm{~Hz}$, and the speed of sound $v$ is $334 \mathrm{~m/s}$.
Using the Doppler effect formula for a source stationary and observer (in this case, the reflector) moving away, the frequency $f'$ heard by the reflector is: $$ f' = f_0 \frac{v}{v + v_r} $$ Where $v_r$ is the velocity of the reflector. Plugging in the values: $$ f' = 334 \frac{334}{334 + 2} = 334 \frac{334}{336} $$
Step 2: Connectivity to reflection
The frequency of the sound wave reflected from the surface will be the same as it received, which is $f'$ (since reflecting surfaces themselves do not change the frequency of sound). Therefore, the frequency of the reflected wave is also $334 \times \frac{334}{336}$.
Step 3: Analyze the second stage as the echo travels back
Now, the echo travels back towards the observer who is stationary relative to the source. Since the wave is approaching the observer, the received frequency $f''$ can be calculated using the Doppler effect formula for observer stationary and source (reflected sound acting as a moving source now) moving towards: $$ f'' = f' \frac{v}{v - v_r} $$ Plugging in $f'$ from step 1 and considering that velocity of reflector is away which is like having source coming closer in equivalence: $$ f'' = \left(334 \times \frac{334}{336}\right) \frac{334}{334 - 2} = 334 \times \frac{334}{336} \times \frac{334}{332} $$
We simplify it by computing: $$ f'' = 334 \frac{334^2}{336 \times 332} = 334 \frac{111556}{111552} $$ This simplifies further to: $$ f'' = 334 \times \frac{13}{12} $$
This value suggests that the apparent frequency $f''$ which observer hears is $334 \times \frac{13}{12}$.
Conclusion
Choosing the correct answer from the options provided:
The factor by which $334 \mathrm{~Hz}$ is multiplied to get the apparent frequency is $\frac{13}{12}$.
So, Option B $\frac{13}{12}$ is correct, representing the frequency multiplier due to the moving reflector and observing conditions.
Find the least number which, when divided by 15, leaves a remainder of 5, when divided by 25, leaves a remainder of 15, and when divided by 35, leaves a remainder of 25.
A) 515
B) 550
C) 530
D) 600
The problem asks us to find the least number that leaves specific remainders when divided by 15, 25, and 35. Let's denote this number as $n$.
The question states:
When $n$ is divided by 15, the remainder is 5. Thus, $n$ can be expressed as: $$ n = 15a + 5 $$ where $a$ is some integer.
When $n$ is divided by 25, the remainder is 15. Hence, $n$ can also be represented as: $$ n = 25b + 15 $$ where $b$ is some integer.
When $n$ is divided by 35, the remainder is 25. Thus, we can write: $$ n = 35c + 25 $$ where $c$ is some integer.
Since these conditions must all be true simultaneously, we start by recognizing that there's a common pattern:
Adding $10$ to $n$ in each case will simplify these expressions to multiples of 15, 25, and 35: $$ \begin{align*} n + 10 &= 15a + 15 \ n + 10 &= 25b + 25 \ n + 10 &= 35c + 35 \end{align*} $$
Notice that now $n + 10$ is a multiple of 15, 25, and 35. We need to find the Least Common Multiple (LCM) of these values. The prime factorizations are:
$15 = 3×5$,
$25 = 5^2$,
$35 = 5×7$.
For the LCM, we take the highest power of each prime number present:
Highest power of 3: $3^1$,
Highest power of 5: $5^2$,
Highest power of 7: $7^1$.
Therefore, $$ LCM(15, 25, 35) = 3^1 × 5^2 × 7^1 = 525 $$
This means $n + 10 = 525m$ for some integer $m$. To find the smallest possible $n$ (i.e., with smallest $m = 1$): $$ n + 10 = 525 \ n = 525 - 10 \ n = 515 $$
Thus, the least number satisfying all these conditions is $515$. Therefore, the answer is Option (A) 515.
The 4th term of an AP is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the first term and the common difference.
Given the problem's constraints, we start with the definition of the $ n^{\text{th}} $ term of an Arithmetic Progression (AP):
$$ a_n = a + (n-1)d $$
where:
( a ) is the first term
( d ) is the common difference
Step 1: Finding the Relationships
4th term is 3 times the first term:
Given: $ a + 3d = 3a $
Rearranging: $ 3d = 2a \quad \ldots \ldots \text{(i)} $
7th term exceeds twice the 3rd term by 1:
Given: $ 2 \times ( 3^{\text{rd}} \text{ term} ) = (7^{\text{th}} \text{ term}) - 1 $
That becomes: $$ 2(a + 2d) = (a + 6d) - 1 $$
Simplifying: $$ 2a + 4d = a + 6d - 1 \ 2a - a + 4d - 6d = -1 \ a - 2d = -1 \quad \ldots \ldots \text{(ii)} $$
Step 2: Solving the Equations
From equation (i): $$ 3d = 2a \implies a = \frac{3d}{2} $$
Substituting $ a = \frac{3d}{2} $ into equation (ii): $$ \frac{3d}{2} - 2d = -1 $$ Multiply the entire equation by 2 to clear the fraction: $$ 3d - 4d = -2 \ -d = -2 \ d = 2 $$
Step 3: Finding the First Term
Substitute $ d = 2 $ back into $ a = \frac{3d}{2} $: $$ a = \frac{3 \times 2}{2} \ a = 3 $$
Conclusion
Therefore:
The first term ( ( a ) ) is 3
The common difference ( ( d ) ) is 2
The sequence is: 3, 5, 7, 9, 11, ...
From the following options, only 46 can be a term of an Arithmetic Progression (AP) whose first term is 2 and the common difference is 4.
23
69
46
92
The correct option is C) 46.
Given:
First term ($a$) = 2
Common difference ($d$) = 4
In an Arithmetic Progression (AP), the $n$-th term is given by: $$ a_n = a + (n-1)d $$ where $a_n$ represents the $n$-th term of the AP, and $n$ must be a natural number.
Substituting the values of $a$ and $d$: $$ a_n = 2 + (n - 1) \cdot 4 = 4n - 2 $$
Let's verify each given option to see if it can be a term of the AP:
When $a_n = 23$: $$ \begin{align*} 4n - 2 &= 23 \ 4n &= 25 \ n &= \frac{25}{4} \end{align*} $$ Here, $n$ is not a natural number.
When $a_n = 46$: $$ \begin{align*} 4n - 2 &= 46 \ 4n &= 48 \ n &= 12 \end{align*} $$ Here, $n = 12$ is a natural number.
When $a_n = 69$: $$ \begin{align*} 4n - 2 &= 69 \ 4n &= 71 \ n &= \frac{71}{4} \end{align*} $$ Here, $n$ is not a natural number.
When $a_n = 92$: $$ \begin{align*} 4n - 2 &= 92 \ 4n &= 94 \ n &= \frac{94}{4} \end{align*} $$ Here, $n$ is not a natural number.
Hence, only 46 is a term of the AP.
Rajeev's age is 6 less than thrice Raju's age. If Raju's age is ( x ) years, then Rajeev's age can be expressed as :
(A) $ 3x-6$
(B) $3x+6$
(C) $x+18 $
(D) $x-18 $
The correct option is A: $3x - 6$.
It is given that Rajeev's age is 6 less than three times Raju's age. Let Raju's age be represented by $ x $ years.
To find Rajeev's age:
First, compute three times Raju's age, which is $ 3x $.
Since Rajeev's age is 6 years less than this value, subtract 6 from $ 3x $.
Therefore, Rajeev's age can be expressed as: $$ 3x - 6 \text{ years} $$
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