Question

In an A.P. let the first term is $a$ and the last is $l$ and the sum of the terms of the A.P. is $S$. Then prove that the common difference is $\frac{(l^2-a^2)}{2S-a-l}$.

Answer 1

Let there are $n$ terms in the A.P. whose first term is $a$ and $n$ th is $l$.

Then $a+(n-1)d=l$

or, $(n-1)d=l-a$

or, $d=\frac{l-a}{n-1}$.....(1).

Again given the sum of the series is $S$.

Then

$S=\frac{n}{2}(a+l)$

or, $n=\frac{2S}{a+l}$.

Now using the value of $n$ in (1) we get,

$d=\frac{(l-a)(l+a)}{2S-l-a}$.

or, $d=\frac{l^2-a^2}{2S-l-a}$.