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Question

For the following two (02) items : 

Let $(6+10+14 + ... \text{up to } m \text{ terms})$ $=(1+3+5+7+ ... \text{up to } n \text{ terms})$ where $m < 25$ and $n < 25$.

How many values of \(m\) are possible?

This question was previously asked in
NDA 2 2025 GAT Question Paper (14-Sep-2025)
The correct answer is
Two

Analyzing the Arithmetic Series

The question asks us to find the number of possible values for '\(m\)' given an equality between the sums of two arithmetic progressions (APs), subject to constraints on '\(m\)' and '\(n\)' (\(m < 25\) and \(n < 25\)).

Let's first find the formula for the sum of each series:

  1. First Series: \(6, 10, 14, ...\) up to \(m\) terms.
    • First term (\(a_1\)) = 6
    • Common difference (\(d_1\)) = \(10 - 6 = 4\)
    • The sum of the first \(m\) terms (\(S_m\)) is given by the formula: \(S_m = \frac{m}{2} [2a_1 + (m-1)d_1]\)
    • Substituting the values: \(S_m = \frac{m}{2} [2(6) + (m-1)4]\) \(S_m = \frac{m}{2} [12 + 4m - 4]\) \(S_m = \frac{m}{2} [8 + 4m]\) \(S_m = m(4 + 2m)\) \(S_m = 2m^2 + 4m\)
  2. Second Series: \(1, 3, 5, 7, ...\) up to \(n\) terms.
    • First term (\(a_2\)) = 1
    • Common difference (\(d_2\)) = \(3 - 1 = 2\)
    • The sum of the first \(n\) terms (\(S_n\)) is given by the formula: \(S_n = \frac{n}{2} [2a_2 + (n-1)d_2]\)
    • Substituting the values: \(S_n = \frac{n}{2} [2(1) + (n-1)2]\) \(S_n = \frac{n}{2} [2 + 2n - 2]\) \(S_n = \frac{n}{2} [2n]\) \(S_n = n^2\)

Setting up the Equality Equation

The problem states that the sums are equal:

\(S_m = S_n\) \(2m^2 + 4m = n^2\)

Finding Possible Values of m

We are given the constraints \(m < 25\) (meaning \(1 \le m \le 24\)) and \(n < 25\) (meaning \(1 \le n \le 24\)). We need to find integer values of \(m\) within its range that result in an integer value of \(n\) within its range, satisfying the equation \(n^2 = 2m^2 + 4m\).

Let's test values of \(m\) from 1 to 24:

m Calculation for \(n^2 = 2m^2 + 4m\) \(n^2\) Is \(n^2\) a perfect square? n Is \(n < 25\)? Valid Pair (m, n)?
1 \(2(1)^2 + 4(1)\) 6 No - - No
2 \(2(2)^2 + 4(2)\) 16 Yes 4 Yes Yes
3 \(2(3)^2 + 4(3)\) 30 No - - No
4 \(2(4)^2 + 4(4)\) 48 No - - No
5 \(2(5)^2 + 4(5)\) 70 No - - No
... ... ... ... ... ... ...
15 \(2(15)^2 + 4(15)\) 510 No - - No
16 \(2(16)^2 + 4(16)\) 576 Yes 24 Yes Yes
17 \(2(17)^2 + 4(17)\) 646 No - - No
... ... ... ... ... ... ...
24 \(2(24)^2 + 4(24)\) 1248 No - - No

By testing values, we find two pairs \((m, n)\) that satisfy the equation and the constraints:

  • When \(m=2\), \(n^2 = 2(2^2) + 4(2) = 8 + 8 = 16\), so \(n=4\). Both \(m=2\) and \(n=4\) are less than 25.
  • When \(m=16\), \(n^2 = 2(16^2) + 4(16) = 2(256) + 64 = 512 + 64 = 576\), so \(n=24\). Both \(m=16\) and \(n=24\) are less than 25.

For any other value of \(m\) between 1 and 24, \(n^2 = 2m^2 + 4m\) does not result in a perfect square for \(n\), or the resulting \(n\) is 25 or greater.

Conclusion on Possible Values of m

The possible values for \(m\) are 2 and 16. Therefore, there are exactly two possible values for \(m\).

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Similar Questions

  1. The fifth term of an AP of n terms, whose sum is n 2– 2n, is

  2. A person is to count 4500 notes. Let a ndenote the number of notes he counts in the nth minute. If a 1= a 2= a 3= … = a 10 = 150, and a 10 , a 11 , a 12 , … are in AP with the common difference -2, then the time taken by him to count all the notes is

  3. If \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) , where S ndenotes the sum of the first n terms of an AP, then the common difference is

  4. If the ratio of AM to GM of two positive numbers a and b is 5 : 3 then a : b is equal to

  5. Let x, y, z be positive real numbers such that x, y, z are in GP and tan -1 x, tan -1 y and tan -1 z are in AP. Then which one of the following is correct?

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  7. If y = x + x 2+ x 3+ … up to infinite terms where x < 1, then which one of the following is correct?

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Important Questions from Sequences and Series

  1. The fifth term of an AP of n terms, whose sum is n 2– 2n, is

  2. A person is to count 4500 notes. Let a ndenote the number of notes he counts in the nth minute. If a 1= a 2= a 3= … = a 10 = 150, and a 10 , a 11 , a 12 , … are in AP with the common difference -2, then the time taken by him to count all the notes is

  3. \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to
  4. If \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) , where S ndenotes the sum of the first n terms of an AP, then the common difference is

  5. What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?

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