Let a₁, a₂, a₃, … be in an A.P. such that Σk=1¹² a₂ₖ₋₁=-72/5 a₁, a₁ ≠ 0. If Σk=1ⁿ aₖ=0, then n is :

Solution

Maths sir · 4 steps

The odd-numbered terms form their own A.P. with difference 2d. Use that to find d in terms of a₁.

1. Odd terms. The terms a₁, a₃, a₅, …, a₂₃ are 12 terms of an A.P. with first term a₁ and common difference 2d. Their sum is 12/2(2a₁ + 11 × 2d) = 12a₁ + 132d.

   12a₁ + 132d = -72/5a₁ ⇒ 132d = -132/5a₁ ⇒ d = -a₁/5

2. Sum equals zero. Sₙ = n/2(2a₁ + (n-1)d) = 0. Since n ≠ 0, we need 2a₁ + (n-1)d = 0, that is 2a₁ - (n-1)a₁/5 = 0.
3. Divide by a1. Because a₁ ≠ 0, divide through by a₁ to get 2 = (n-1)/5. This gives n - 1 = 10.
4. So. n = 11, which is option (c).

The idea to keep. Picking every second term of an A.P. gives another A.P. with difference 2d. Turn every condition into a relation between a₁ and d, and a₁ will cancel at the end.

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