Let the positive numbers a₁, a₂, a₃, a₄ and a₅ be in a G.P. Let their mean and variance be 31/10 and m/n respectively, where m and n are co-prime. If the mean of their reciprocals is 31/40 and a₃+a₄+a₅=14, then m+n is equal to ___________.

Solution

Maths sir · 4 steps

Divide the sum by the sum of reciprocals: in a 5-term G.P. that ratio is just a₃².

1. Pair the terms. In a G.P. of 5 terms, a₁ a₅ = a₂ a₄ = a₃². So 1/aᵢ = a₆₋ᵢ/a₃², and the sum of reciprocals equals S/a₃², where S is the sum of the terms.

   a₃² = (Σ aᵢ)/(Σ 1/aᵢ) = (5 × 31/10)/(5 × 31/40) = (31/2)/(31/8) = 4

2. Find the ratio. The terms are positive, so a₃ = 2. Then 2 + 2r + 2r² = 14 gives r² + r - 6 = 0, so r = 2 (the root r = -3 would make some terms negative). The terms are 1/2, 1, 2, 4, 8, and their sum is 15.5, which matches.
3. Variance. The sum of squares is 1/4 + 1 + 4 + 16 + 64 = 341/4, so the mean of the squares is 341/20 = 1705/100. Subtract the square of the mean: 1705/100 - 961/100 = 744/100 = 186/25.
4. So. m = 186 and n = 25 are co-prime, so m + n = 211.

The common slip: 844. Leaving the variance as 744/100 without cancelling the common factor 4 gives 844. Reduce to 186/25 first, because m and n must be co-prime.

The idea to keep. In a G.P. with an odd number of terms, terms equally far from the middle multiply to the middle term squared. So the sum divided by the sum of reciprocals gives that square directly.

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