Let a₁, a₂, a₃, … be an A.P. If Σᵣ₌₁^∞ aᵣ/2ʳ=4, then 4 a₂ is equal to _________.

Solution

Maths sir · 4 steps

This is an AGP: halve the sum, subtract, and you are left with a plain GP.

1. Write the sum. Let aᵣ = a₁ + (r-1)d and S = a₁/2 + a₂/4 + a₃/8 + … = 4. Then S/2 = a₁/4 + a₂/8 + a₃/16 + ….
2. Subtract. Subtract term by term, lining up equal powers of 2. Each difference aᵣ₊₁ - aᵣ is d, so after the first term only a geometric series remains.

   S - S/2 = a₁/2 + d(1/4 + 1/8 + …) = a₁/2 + d/2

3. Find a2. The GP sums to (1/4)/(1 - 1/2) = 1/2. So S/2 = (a₁ + d)/2, which gives S = a₁ + d = a₂. Since S = 4, a₂ = 4.
4. So. 4a₂ = 4 × 4 = 16.

The common slip: 4. Stopping at a₂ = 4 gives 4. The question asks for 4a₂, which is 16.

The idea to keep. For a sum of (AP term) times (GP term), multiply by the common ratio and subtract. The AP differences become a constant d, and what is left is an ordinary GP.

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