NDA Mathematics Question (2025) — Solution

Step-by-step solution

The sum of the first m terms of the arithmetic progression 6, 10, 14, is given by: S_m = m 2 [2(6) + (m - 1)4] = m 2 [12 + 4m - 4] = 2m^2 + 4m The sum of the first n terms of the arithmetic progression 1, 3, 5, 7, is given by: S_n = n 2 [2(1) + (n - 1)2] = n^2 Equating the two sums, we get: 2m^2 + 4m = n^2 2(m^2 + 2m + 1) - 2 = n^2 2(m + 1)^2 - n^2 = 2 Let u = m + 1. The equation becomes 2u^2 - n^2 = 2. This implies that n must be an even number. Let n = 2v. Substituting this into the equation yields: 2u^2 - 4v^2 = 2 u^2 - 2v^2 = 1 This is a standard Pell's equation. The fundamental solution is u_1 = 3 and v_1 = 2. For u_1 = 3 and v_1 = 2, we have m + 1 = 3 m = 2 and n = 2(2) = 4. Both m The next solution is generated by (u_1 + v_1 2 )^2 = (3 + 2 2 )^2 = 17 + 12 2 . Thus, u_2 = 17 and v_2 = 12, which gives m + 1 = 17 m = 16 and n = 2(12) = 24. Both m The third solution is generated by (3 + 2 2 )^3 = 99 + 70 2 . Thus, u_3 = 99 and v_3 = 70, which gives m + 1 = 99 m = 98 and n = 2(70) = 140. This does not satisfy the given conditions m Therefore, the only possible values for m are 2 and 16. There are exactly two possible values. Answer: Two