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 given by $\frac{l^2-a^2}{k-(l+a)}$, then k =
(a) S
(b) 2S
(c) 3S
(d) none of these

Answer 1

(b) 2S

Given:

$$\begin{gathered} S=\frac{n}{2}\left(l+a\right) \\ \Rightarrow \left(l+a\right)=\frac{2S}{n} \end{gathered}$$

$$\begin{gathered} \mathrm{Also},d=\frac{l^2-a^2}{k-\left(l+a\right)} \\ \Rightarrow d=\frac{\left(l+a\right)\left(l-a\right)}{k-\left(l+a\right)} \\ \Rightarrow d=\frac{\left[\left(n-1\right)d\right]\times {\displaystyle \frac{2S}{n}}}{k-{\displaystyle \frac{2S}{n}}} \\ \Rightarrow k-\frac{2S}{n}=\left(n-1\right)\frac{2S}{n} \\ \Rightarrow k=\frac{2S}{n}\left(n-1+1\right) \\ \Rightarrow k=2S \end{gathered}$$