If the sum of the series 1/(1 ·(1+d))+1/(1+d)(1+2 d)+…+1/(1+9 d)(1+10 d) is equal to 5, then 50 d is equal to :

Solution

Maths sir · 4 steps

Consecutive terms of an AP in the denominator: split each fraction and the middle terms cancel.

1. Split each term. The denominators are neighbouring terms of an AP with common difference d. Their difference is d, so each fraction breaks into a difference of two simpler fractions.

   1/(1+kd)(1+(k+1)d) = 1/d(1/(1+kd)-1/(1+(k+1)d))

2. Telescope. Add this from k = 0 to k = 9. Every middle term cancels with the next one, leaving only 1/d(1-1/(1+10d)) = 1/d × 10d/(1+10d) = 10/(1+10d).
3. Use the sum. Set 10/(1+10d) = 5. Then 1+10d = 2, so d = 1/10.
4. So. 50d = 50 × 1/10 = 5, which is option (a).

The idea to keep. When a denominator is a product of consecutive AP terms, write 1/a(a+d) = 1/d(1/a-1/(a+d)). Only the first and last pieces survive.

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