If S = 7/5 + 9/5² + 13/5³ + 19/5⁴ + ...., then 160 S is equal to ________.

Solution

Maths sir · 4 steps

Numerators differ by 2, 4, 6, 8, so subtract S/5 twice and the series becomes a plain GP.

1. Spot the pattern. The numerators 7, 9, 13, 19, … have differences 2, 4, 6, …, which form an AP. So one shift-and-subtract will not finish the job. You need it twice.
2. First subtraction. Write S/5 = 7/5² + 9/5³ + 13/5⁴ + … and subtract it from S term by term. This gives 4/5S = 7/5 + T, where T = 2/5² + 4/5³ + 6/5⁴ + …
3. Second subtraction. Do the same to T: T - T/5 = 2/5² + 2/5³ + 2/5⁴ + …, an infinite GP with ratio 1/5. Its sum is (2/25)/(4/5) = 1/10, so 4/5T = 1/10 and T = 1/8.

   4/5S = 7/5 + 1/8 = 61/40 ⇒ S = 61/32

4. So. 160S = 160 × 61/32 = 5 × 61 = 305.

The common slip: 244. Stopping at 4/5S = 61/40 and taking S = 61/40 gives 160S = 244. You still have to divide by 4/5.

The idea to keep. If the numerators have differences in AP, multiply by the ratio and subtract twice. Each subtraction knocks the numerator down one level until a plain GP is left.

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