Question

The ratio of the sum of the first \(m\) terms to the sum of the first \(n\) terms of an arithmetic progression is \(m^2 : n^2\). What is the ratio of its 17th term to the 29th term?

• A. 29 : 41
• B. 9 : 17
• C. 13 : 21
• D. 11 : 19

Hint:

Use the nth term formula \(T_n = a + (n-1)d\) to solve for the ratio of terms. This can be helpful when the ratio of sums is given as a square.

Answer 1

The Correct Option is A

We are given that the ratio of the sums of the first \(m\) and \(n\) terms is \(m^2 : n^2\), i.e., \[ \frac{S_m}{S_n} = \frac{m^2}{n^2} \] The sum of the first \(n\) terms of an arithmetic progression is: \[ S_n = \frac{n}{2} \left(2a + (n-1) d\right) \] Thus, the ratio of sums \(S_m\) and \(S_n\) can be written as: \[ \frac{S_m}{S_n} = \frac{\frac{m}{2} [2a + (m-1) d]}{\frac{n}{2} [2a + (n-1) d]} \] Simplifying, we get: \[ \frac{S_m}{S_n} = \frac{m [2a + (m-1) d]}{n [2a + (n-1) d]} \] Since the ratio is \(m^2 : n^2\), we deduce the common difference and first term follow this pattern. Thus, the ratio of the 17th term to the 29th term is: \[ \frac{T_{17}}{T_{29}} = \frac{17}{29} \] Therefore, the ratio of the 17th to the 29th term is 29:41.