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 $\Large\frac{l^{2}-a^{2}}{k-(l+a)}$ then k = ?

A. S B. 2S C. 3S D. None of these Answer: Option B
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Solution(By Apex Team)

$\begin{aligned}S&=\frac{n}{2}(l+a)\\ l&=a+(n-1)d\\ d&=\frac{l^2-a^2}{k-(l+a)}\text{ and also }d=\frac{l-a}{n-1}\\ &\therefore\frac{l-a}{n-1}=\frac{(l+a)(l-a)}{k-(l+a)}\\ &\Rightarrow\frac{1}{n-1}=\frac{l+a}{k-(l+a)}\\ &\Rightarrow k-(l+a)=(n-1)(l+a)\\ &\Rightarrow k=(n-1)(l+a)+(l+a)\\ &\Rightarrow k=(l+a)(n-1+1)\\ &\Rightarrow k=n(l+a)\\ &\Rightarrow k=2\times\frac{n}{2}(l+a)\left\{\therefore\frac{n}{2}(l+a)=S\right\}\\ &\Rightarrow k=2\times S\\ &\Rightarrow k=2S\end{aligned}$

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