Sₙ(x) = log a 1/2 x + log a 1/3 x + log a 1/6 x + log a 1/11 x + log a 1/18 x + log a 1/27 x + ...... up to n-terms, where a > 1. If S₂₄(x) = 1093 and S₁₂(2x) = 265, then value of a is equal to ____________.

Solution

Maths sir · 4 steps

Pull the root out of each base as a multiplier, and the series becomes a sum of numbers times logₐ x.

1. Fix each base. Since log_(a^(1/k)) x = k logₐ x, every term is a plain number times logₐ x. The multipliers are 2, 3, 6, 11, 18, 27, …
2. Find the pattern. The differences are 1, 3, 5, 7, 9, …, the odd numbers. So the rth multiplier is tᵣ = (r-1)² + 2.

   Σᵣ₌₁ⁿ tᵣ = (n-1)n(2n-1)/6 + 2n

3. Use both conditions. For n = 24: (23 × 24 × 47)/6 + 48 = 4324 + 48 = 4372, so 4372 logₐ x = 1093 and logₐ x = 1/4. For n = 12: (11 × 12 × 23)/6 + 24 = 506 + 24 = 530, so 530 logₐ (2x) = 265 and logₐ(2x) = 1/2.
4. So. logₐ 2 = logₐ(2x) - logₐ x = 1/2 - 1/4 = 1/4, which gives a = 2⁴ = 16.

The idea to keep. When a base carries a power, log_(aᵖ) x = 1/p logₐ x turns a messy log series into an ordinary number series. Then check the differences of the coefficients to find the general term.

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