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 :

Step-by-step solution

aligned & 1 1 (1+ d ) + 1 (1+ d )(1+2 ~d ) + . . & 1 (1+9 ~d )(1+10 ~d ) =5 aligned aligned & 1 d [ (1+d)-1 1 (1+d) + (1+2 d)-(1-d) (1+d)(1+2 d) ]+ . . . & (1+10 d)-(1+9 d) (1+9 d)(1+10 d) =5 & 1 d [ (1- 1 1+d )+ ( 1 1+d - 1 1+2 d )+ . . . & . ( 1 1+9 d - 1 1+10 d ) ]=5 & 1 d [1- 1 (1+10 d) ]=5 & 10 d 1+10 d =5 d & 50 d=5 aligned