Question:
The first and last term of an A.P. are a and $/$ respectively. If $S$ is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^{2}-a^{2}}{k-(l+a)}$, then $k=$
(a) S
(b) 2S
(c) 3S
(d) none of these
Solution:
(b) 2S
Given:
$S=\frac{n}{2}(l+a)$
$\Rightarrow(l+a)=\frac{2 S}{n}$
Also, $d=\frac{l^{2}-a^{2}}{k-(l+a)}$
$\Rightarrow d=\frac{(l+a)(l-a)}{k-(l+a)}$
$\Rightarrow d=\frac{[(n-1) d] \times \frac{2 S}{n}}{k-\frac{25}{n}}$
$\Rightarrow k-\frac{2 S}{n}=(n-1) \frac{2 S}{n}$
$\Rightarrow k=\frac{2 S}{n}(n-1+1)$
$\Rightarrow k=2 S$
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