If Sₙ=4+11+21+34+50+… to n terms, then 1/60( S₂₉-S₉) is equal to :

Solution

Maths sir · 4 steps

The gaps 7, 10, 13, 16 form an AP, so the nth term is a quadratic in n. Find it, then sum it.

1. Look at differences. The gaps between terms are 7, 10, 13, 16, …, an AP with common difference 3. Constant second differences mean Tₙ is a quadratic in n, of the form Tₙ = an² + bn + c with 2a = 3.
2. Find the general term. With a = 3/2, putting in T₁ = 4 and T₂ = 11 gives b + c = 5/2 and 2b + c = 5. So b = 5/2 and c = 0, which gives Tₙ = n(3n+5)/2. Check: T₃ = (3 × 14)/2 = 21 and T₄ = (4 × 17)/2 = 34.

   Sₙ = 1/2[3 · n(n+1)(2n+1)/6 + 5 · n(n+1)/2] = n(n+1)(n+3)/2

3. Put in numbers. S₂₉ = (29 × 30 × 32)/2 = 13920 and S₉ = (9 × 10 × 12)/2 = 540. Their difference is 13380.
4. So. 1/60(S₂₉ - S₉) = 13380/60 = 223, which is option (d).

The idea to keep. When the differences of a series form an AP, the nth term is quadratic in n. Fit it with the first two or three terms, then use Σ n² and Σ n. Always check S₁ and S₂ at the end.

Answer from the NTA final answer key. Solution by PrepWiser.