Let aₙ be the nᵗʰ term of the series 5+8+14+23+35+50+… and Sₙ=Σk=1ⁿ aₖ. Then S₃₀-a₄₀ is equal to :

Solution

Maths sir · 4 steps

The gaps are 3, 6, 9, 12, … so write each term as 5 plus three times a triangular number.

1. Look at differences. The differences between consecutive terms are 3, 6, 9, 12, 15, …, that is 3 × 1, 3 × 2, 3 × 3, …. To reach aₙ from a₁ = 5 you add the first n-1 of them.

   aₙ = 5 + 3(1 + 2 + … + (n-1)) = 5 + 3n(n-1)/2

2. Find the 40th term. Check: a₆ = 5 + (3 × 6 × 5)/2 = 50, which matches. Then a₄₀ = 5 + (3 × 40 × 39)/2 = 5 + 2340 = 2345.
3. Sum 30 terms. S₃₀ = 5 × 30 + 3/2Σₖ₌₁³⁰ k(k-1). Here Σ k² = (30 × 31 × 61)/6 = 9455 and Σ k = 465, so Σ k(k-1) = 8990. That gives S₃₀ = 150 + 3/2 × 8990 = 150 + 13485 = 13635.
4. So. S₃₀ - a₄₀ = 13635 - 2345 = 11290, which is option (b).

The idea to keep. If the differences of a series form an AP, the nth term is a quadratic in n. Find it from the sum of the first n − 1 differences, then sum it with the formulas for Σk and Σk².

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