NDA Mathematics Question (2025) — Solution

Step-by-step solution

The sum of the first k terms of the series is given by S_k = 3k^2 + 5k. The k-th term of the series, T_k, is obtained by subtracting the sum of the first k-1 terms from the sum of the first k terms: T_k = S_k - S_ k-1 Substituting the given expression for S_k and S_ k-1 : T_k = (3k^2 + 5k) - [3(k-1)^2 + 5(k-1)] T_k = 3k^2 + 5k - [3(k^2 - 2k + 1) + 5k - 5] T_k = 3k^2 + 5k - (3k^2 - 6k + 3 + 5k - 5) T_k = 3k^2 + 5k - (3k^2 - k - 2) T_k = 6k + 2 To find the common difference, we calculate the difference between consecutive terms: T_k - T_ k-1 = (6k + 2) - [6(k-1) + 2] T_k - T_ k-1 = 6k + 2 - (6k - 6 + 2) T_k - T_ k-1 = 6 Since the difference between consecutive terms is a constant, the terms of the series form an arithmetic progression with a common difference of 6. Answer: The terms of S form an arithmetic progression with common difference 6.