Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be 98/25. Then the sum of the first 21 terms of an AP, whose first term is 10a r, nᵗʰ term is aₙ and the common difference is 10 ar², is equal to :

Solution

Maths sir · 3 steps

An AP with an odd number of terms has a middle term, and the sum is the count times that middle term.

1. GP data. From a/(1-r) = 5 and a(1-r⁵)/(1-r) = 98/25 we get 1 - r⁵ = 98/125, so r⁵ = 27/125. So r ≠ 0 and a = 5(1-r) ≠ 0. The AP is a genuine one with first term 10ar and common difference 10ar².
2. Sum of 21 terms. For any AP, S₂₁ = 21/2(a₁ + a₂₁). Write a₁ = a₁₁ - 10d and a₂₁ = a₁₁ + 10d. The two 10d terms cancel.

   S₂₁ = 21/2(2a₁₁) = 21 a₁₁

3. So. The sum of the first 21 terms is 21 a₁₁, which is option (a). The actual values of a and r are not needed.

The idea to keep. For an AP with an odd number n of terms, Sₙ = n × (middle term). The middle term is a_((n+1)/2). Spot this before you start solving for a and r.

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