Let a and b be be two distinct positive real numbers. Let 11ᵗʰ term of a GP, whose first term is a and third term is b, is equal to pᵗʰ term of another GP, whose first term is a and fifth term is b. Then p is equal to

Solution

Maths sir · 4 steps

Write both terms as powers of the same base b/a, then match the powers.

1. First GP. With ratio r, the third term gives ar² = b, so r² = b/a. Then T₁₁ = ar¹⁰ = a(r²)⁵ = a(b/a)⁵.
2. Second GP. With ratio s, the fifth term gives as⁴ = b, so s⁴ = b/a and s = (b/a)^(1/4). Then Tₚ = asᵖ⁻¹ = a(b/a)^((p-1)/4).

   a(b/a)⁵ = a(b/a)^((p-1)/4)

3. Match the powers. Since a ≠ b, the base b/a is a positive number other than 1, so equal values force equal exponents: (p-1)/4 = 5.
4. So. p - 1 = 20, which gives p = 21, option (c).

The idea to keep. When two GPs share a first term, express every term as a times a power of the same ratio b/a. Then compare exponents, and remember the nth term carries rⁿ⁻¹.

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