Let { aₖ} and { bₖ} ,k ∈ N, be two G.P.s with common ratios r₁ and r₂ respectively such that a₁ = b₁ = 4 and r₁ < r₂. Let cₖ = aₖ + bₖ,k ∈ N. If c₂ = 5 and c₃ = 13/4 then Σₖ₌₁^∞ cₖ - (12a₆ + 8b₄) is equal to __________.

Solution

Maths sir · 4 steps

Both G.P.s start at 4, so c₂ and c₃ give you the sum and product of r₁, r₂: build the quadratic.

1. Use the terms. c₂ = 4r₁ + 4r₂ = 5 gives r₁ + r₂ = 5/4. c₃ = 4r₁² + 4r₂² = 13/4 gives r₁² + r₂² = 13/16.
2. Find the product. 2r₁r₂ = (r₁+r₂)² - (r₁²+r₂²) = 25/16 - 13/16 = 12/16, so r₁r₂ = 3/8. The ratios are the roots of 8t² - 10t + 3 = 0, which are t = 1/2 and t = 3/4. Since r₁ < r₂, r₁ = 1/2 and r₂ = 3/4.

   Σₖ₌₁^∞ cₖ = 4/(1-1/2) + 4/(1-3/4) = 8 + 16 = 24

3. The subtracted part. a₆ = 4(1/2)⁵ = 1/8, so 12a₆ = 3/2. b₄ = 4(3/4)³ = 27/16, so 8b₄ = 27/2. Together they make 3/2 + 27/2 = 15.
4. So. The required value is 24 - 15 = 9.

The idea to keep. When two unknown ratios appear symmetrically, find their sum and product and solve one quadratic. Then use the given order (r₁ < r₂) to decide which root belongs to which G.P.

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