Let a₁, a₂, a₃, … be a G.P. of increasing positive terms. If a₁ a₅=28 and a₂+a₄=29, then a₆ is equal to:

Solution

Maths sir · 4 steps

In a G.P., terms equally far from the middle have the same product, so a₁a₅ is also a₂a₄.

1. Pair the terms. With first term a and ratio r, a₁a₅ = a · ar⁴ = a²r⁴ and a₂a₄ = ar · ar³ = a²r⁴. So a₂a₄ = 28 as well.
2. Sum and product. Now a₂ and a₄ have sum 29 and product 28, so they are the roots of t² - 29t + 28 = 0, which factorises as (t-1)(t-28) = 0. The terms increase, so a₂ = 1 and a₄ = 28.
3. Find the ratio. Going from a₂ to a₄ multiplies by r², so r² = 28/1 = 28.

   a₆ = a₄ · r² = 28 × 28 = 784

4. So. a₆ = 784, which is option (b).

The idea to keep. In a G.P., when the subscripts add to the same number, the products match: a₁a₅ = a₂a₄ = a₃². Turn a given sum and product into a quadratic and read off the two terms.

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