If the sum of the series 11 (1 + d) + 1(1 + d)(1 + 2d) + + 1(1 + 9d)(1 + 10d) is equal to 5, then 50d is equal to :
- 5
- 10
- 15
- 20
Answer
(1) 5
Solution
The denominators are products of consecutive terms of an A.P. with common difference d, so each term splits as
1(1 + rd)(1 + (r + 1)d) = 1d [ 11 + rd - 11 + (r + 1)d ]
Adding, the middle terms cancel (telescoping):
Sum = 1d [1 - 11 + 10d ] = 101 + 10d = 5
1 + 10d = 2 d = 110, hence 50d = 5
Answer: (1)