Let a₁=8, a₂, a₃, …, aₙ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is ___________.

Solution

Maths sir · 4 steps

Get d from the first four terms, then pair them with the last four to find the last term and n.

1. Find d. The first four terms add to 4a₁ + 6d = 32 + 6d. Setting this equal to 50 gives 6d = 18, so d = 3.
2. Pair the ends. In an A.P., terms equally far from the two ends add to the same value: a₁ + aₙ = a₂ + aₙ₋₁ = a₃ + aₙ₋₂ = a₄ + aₙ₋₃. So the first four and last four terms together make four such pairs.

   4(a₁ + aₙ) = 50 + 170 = 220 ⇒ aₙ = 55 - 8 = 47

3. Count the terms. From aₙ = 8 + (n-1) × 3 = 47 you get n - 1 = 13, so n = 14. With 14 terms, the middle two are a₇ and a₈.
4. So. a₇ = 8 + 6 × 3 = 26 and a₈ = 8 + 7 × 3 = 29. Their product is 26 × 29 = 754.

The idea to keep. In an A.P., the first and last terms, the second and second-last, and so on, all have the same sum. Use this whenever a question gives you sums from both ends.

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