Let a₁, a₂, ..........., a₂₁ be an AP such that Σₙ₌₁²⁰ 1/aₙaₙ₊₁ = 4/9. If the sum of this AP is 189, then a₆a₁₆ is equal to :

Solution

Maths sir · 4 steps

Split each term as a difference of reciprocals and the 20-term sum telescopes to just two terms.

1. Telescope the sum. With common difference d, each term splits as 1/(aₙ aₙ₊₁) = 1/d(1/aₙ - 1/aₙ₊₁). Adding all 20 terms, the middle ones cancel and only the first and last survive.

   Σₙ₌₁²⁰ 1/(aₙ aₙ₊₁) = 1/d(1/a₁ - 1/a₂₁) = (a₂₁ - a₁)/(d a₁ a₂₁) = 20d/(d a₁ a₂₁) = 20/(a₁ a₂₁)

2. Use both conditions. 20/(a₁ a₂₁) = 4/9 gives a₁ a₂₁ = 45. The sum of 21 terms is 21/2(a₁ + a₂₁) = 189, so a₁ + a₂₁ = 18. Then a₁ and a₂₁ are 3 and 15, so 20d = ± 12, giving d = ± 0.6, and the middle term is a₁₁ = 18/2 = 9.
3. Pair around the middle. a₆ and a₁₆ sit 5 places on either side of a₁₁, so a₆ a₁₆ = (a₁₁ - 5d)(a₁₁ + 5d) = a₁₁² - 25d². The sign of d does not matter.
4. So. a₆ a₁₆ = 81 - 25 × 0.36 = 81 - 9 = 72, which is option (b).

The idea to keep. A sum of 1/(aₙ aₙ₊₁) over an AP always telescopes to n/(a₁ aₙ₊₁). Terms placed symmetrically about the middle term multiply to (middle)² - (kd)².

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