If for x, y ∈ R, x > 0, y = log₁₀x + log₁₀x 1/3 + log₁₀x 1/9 + ...... upto ∞ terms
and (2 + 4 + 6 + .... + 2y)/(3 + 6 + 9 + ..... + 3y) = 4/(log ₁₀x), then the ordered pair (x, y) is equal to :

Solution

Maths sir · 4 steps

Pull log₁₀ x out of every term, and the rest is an infinite GP with ratio 1/3.

1. Sum the GP. Each term is log₁₀ xᵏ = k log₁₀ x, so y = log₁₀ x(1 + 1/3 + 1/9 + …). The bracket is an infinite GP with first term 1 and ratio 1/3, so it sums to 1/(1 - 1/3) = 3/2.

   y = 3/2 log₁₀ x

2. Simplify the ratio. The top is 2(1 + 2 + … + y) and the bottom is 3(1 + 2 + … + y). The common sum cancels, so the left side is just 2/3.
3. Find log x. 2/3 = 4/(log₁₀ x) gives log₁₀ x = 6, so x = 10⁶. Then y = 3/2 × 6 = 9.
4. So. The ordered pair is (10⁶, 9), which is option (d).

The idea to keep. When every term is a log of a power of the same base, pull the log out first. What is left is usually a plain GP you can sum with a/(1-r).

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