Let S=109+108/5+107/5²+… .+2/5¹⁰⁷+1/5¹⁰⁸. Then the value of (16 S-(25)⁻⁵⁴) is equal to ___________.

Solution

Maths sir · 4 steps

Numerators drop by 1 while powers of 5 rise: that is an AGP, so subtract S/5 from S.

1. Spot the AGP. The numerators 109, 108, …, 1 form an AP, and the factors 1, 1/5, 1/5², … form a GP with ratio 1/5. Write S/5 and line it up under S, shifted one place to the right.

   S - S/5 = 109 - (1/5 + 1/5² + … + 1/5¹⁰⁸) - 1/5¹⁰⁹

2. Sum the GP. Each matching pair of terms differs by -1 times a power of 1/5. The bracket has 108 terms, so it equals (1/5(1 - 5⁻¹⁰⁸))/(1 - 1/5) = (1 - 5⁻¹⁰⁸)/4.
3. Multiply by 20. So 4S/5 = 109 - 1/4 + 5⁻¹⁰⁸/4 - 5⁻¹⁰⁹. Multiplying by 20 gives 16S = 2180 - 5 + 5⁻¹⁰⁷ - 4 × 5⁻¹⁰⁸ = 2175 + 5⁻¹⁰⁸(5 - 4) = 2175 + 5⁻¹⁰⁸.
4. So. Since (25)⁻⁵⁴ = 5⁻¹⁰⁸, the small leftover term cancels exactly: 16S - (25)⁻⁵⁴ = 2175.

The idea to keep. For any AP-times-GP series, multiply by the common ratio and subtract. The leftover is a plain GP, and the stray tiny powers are usually what the question asks you to remove.

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