If the sum of an infinite GP a, ar, ar², ar³, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar², ar⁴, ar⁶, ....... is :

Solution

Maths sir · 5 steps

Divide the sum of squares by the square of the sum: a cancels and r drops out in one line.

1. Two equations. The GP sums to a/(1-r) = 15. The squares a², a²r², a²r⁴, … form a GP with ratio r², so a²/(1-r²) = 150.
2. Eliminate a. Divide the second equation by the square of the first. Using 1-r² = (1-r)(1+r), one factor of 1-r cancels.

   (a²/(1-r²))/(a²/(1-r)²) = (1-r)/(1+r) = 150/225 = 2/3

3. Find r and a. 3 - 3r = 2 + 2r gives r = 1/5. Then a = 15(1-r) = 15 × 4/5 = 12.
4. Required sum. The terms ar², ar⁴, ar⁶, … form a GP with first term ar² = 12/25 and ratio r² = 1/25. Its sum is ar²/(1-r²) = (12/25)/(24/25) = 12/24.
5. So. The sum is 1/2, which is option (b).

The idea to keep. When you know the sum and the sum of squares of an infinite GP, divide the second by the square of the first. You get (1-r)/(1+r), so r comes out directly.

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