Let the first three terms 2, p and q, with q ≠ 2, of a G.P. be respectively the 7ᵗʰ, 8ᵗʰ and 13ᵗʰ terms of an A.P. If the 5ᵗʰ term of the G.P. is the nᵗʰ term of the A.P., then n is equal to:

Solution

Maths sir · 4 steps

Write p and q using the A.P. difference d, then the G.P. condition gives d.

1. Use the A.P. The 7th term is 2. The 8th term is one step later and the 13th is six steps later, so p = 2 + d and q = 2 + 6d.

   p² = 2q ⇒ (2+d)² = 2(2+6d) ⇒ d² = 8d

2. Find d. d = 0 would make q = 2, which is not allowed, so d = 8. Then p = 10 and q = 50, and the G.P. has ratio r = 5.
3. Fifth G.P. term. T₅ = 2 × 5⁴ = 1250. Counting from the 7th A.P. term, 1250 = 2 + (n-7) × 8, so n - 7 = 156.
4. So. n = 163, which is option (c).

The idea to keep. When terms of one progression sit inside another, express them all through the A.P. difference d. One G.P. condition then fixes d.

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