Question

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 $\frac{l^2-a^2}{k-(l+a)}$, then $k$ is equal to

• A. $S$
• B. $2S$
• C. $3S$
• D. None of these

Answer 1

Answer: (B)

$2S$

Last term, $l=a+(n-1)d$
$d=$ $\frac{l^2-a^2}{k-(l+a)}$
$l=a+(n-1)\times \frac{l^2-a^2}{k-(l+a)}$
$l-a=(n-1)\times \frac{l^2-a^2}{k-(l+a)}$
$\frac{k-(l+a)}{l+a}+1=n$
$n=\frac{k}{l+a}$
Sum of $n$ terms, $S=$ $\frac{n}{2}\left(2a+(n-1)d\right)$
$S=\frac{k}{2l+2a}\times (2a+(\frac{k}{l+a}-1)\times \frac{l^2-a^2}{k-(l+a)})$
$S=$ $\frac{k}{2l+2a}\times (2a+\left(\frac{k-l-a}{l+a}\right)\times \frac{l^2-a^2}{k-(l+a)})$
$S=$ $\frac{k}{2l+2a}\times (2a+l-a)$
$S=$ $\frac{k}{2l+2a}\times (a+l)$
$S=$ $\frac{k}{2}$
$k=2S$