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

Solution(By Examveda Team)

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