Let Sₙ be the sum to n-terms of an arithmetic progression 3,7,11, If 40<(6/n(n+1) Σk=1ⁿ Sₖ)<42, then n equals ________.

Solution

Maths sir · 4 steps

Write Sₙ as a quadratic in n, sum it, and the whole bracket shrinks to just 4n + 5.

1. Sum of n terms. Here a = 3 and d = 4. So Sₙ = n/2(6 + 4(n-1)) = n(2n+1) = 2n² + n.
2. Add up the sums. Use Σ k² = n(n+1)(2n+1)/6 and Σ k = n(n+1)/2, then take out n(n+1) as a common factor.

   Σₖ₌₁ⁿ Sₖ = n(n+1)(2n+1)/3 + n(n+1)/2 = n(n+1)(4n+5)/6

3. Simplify the bracket. Multiplying by 6/n(n+1) leaves only 4n + 5. The condition becomes 40 < 4n + 5 < 42, that is 35 < 4n < 37.
4. So. The only multiple of 4 between 35 and 37 is 36, so 4n = 36 and n = 9.

The idea to keep. To sum the partial sums of an AP, write Sₖ as a quadratic in k and use the formulas for Σ k² and Σ k. Always factor out n(n+1) early.

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