The first and last term of an A.P. are `a` and `l` respectively. If `S` is the sum of all the terms of the A.P. and the common difference is given by `(l^2-a^2)/(k-(l+a))` , then `k=` (a)`S `(b) `2S` (c) `3S` (d) none of these

Explanation

We have, `S=n/2​(a+l)⇒(2S)/(a+l)​=n ...(1)`


Also,` l=a+(n−1)d`

`=>d=(l−a​)/(n−1)`

`=(l−a​ )/((2S​)/(a+l)−1)` (from (1)

`=((l^2−a^2)​)/((2S−(l+a)))`


`k=2S`