Question
For the following two (02) items: Let (6 + 10 + 14 + up to m terms ) = (1 + 3 + 5 + 7 + up to n terms ), where m What is the relation between m and n?
NDA
Mathematics
Sequences and Series
2025
NDA 2025 (Phase 2)
NDA Mathematics Question (2025) — Solution
Options
- A. n^2 = m(m+1)
- B. n^2 = m(m+2)
- C. n^2 = 2m(m+1)
- D. n^2 = 2m(m+2)
Answer
D. n^2 = 2m(m+2)
Step-by-step solution
The first series is an arithmetic progression with first term a = 6 and common difference d = 4. The sum up to m terms is given by: S_m = m 2 [2(6) + (m-1)4] S_m = m 2 [12 + 4m - 4] = m 2 [4m + 8] = 2m(m + 2) The second series is an arithmetic progression with first term a = 1 and common difference d = 2 (which is the sum of the first n odd natural numbers). The sum up to n terms is given by: S_n = n 2 [2(1) + (n-1)2] = n^2 Equating the two sums, we get: n^2 = 2m(m + 2)
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