Let aₙ be the nᵗʰ term of a G.P. of positive terms.
Σₙ₌₁¹⁰⁰ a₂ₙ₊₁ = 200 and Σₙ₌₁¹⁰⁰ a₂ₙ = 100,
then Σₙ₌₁²⁰⁰ aₙ is equal to :

Solution

Maths sir · 4 steps

Line up the two sums term by term: each odd-indexed term is r times the even one before it, so r = 2.

1. Write out both sums. The first sum is a₃ + a₅ + … + a₂₀₁ = 200. The second is a₂ + a₄ + … + a₂₀₀ = 100. Both have 100 terms.
2. Find the ratio. Pair them term by term: a₃ = r a₂, a₅ = r a₄, and so on up to a₂₀₁ = r a₂₀₀. So the first sum is r times the second, which gives 200 = 100r and r = 2.
3. Missing odd sum. We still need a₁ + a₃ + … + a₁₉₉. Each even term is r times the odd term just before it: a₂ = r a₁, a₄ = r a₃, and so on. So this odd sum is 100/r = 100/2 = 50.

   Σₙ₌₁²⁰⁰ aₙ = (a₁ + a₃ + … + a₁₉₉)₅₀ + (a₂ + a₄ + … + a₂₀₀)₁₀₀

4. So. The first 200 terms add up to 50 + 100 = 150, which is option (a).

The idea to keep. When a GP is split into alternate terms, compare the two sums term by term. Their ratio is a power of r. Check the index range carefully before you add them.

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