Let S denotes the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n – k S_n-1 + S_n-2 then k =

Solution(By Examveda Team)

S_n is the sum of n terms of an A.P.
a is its first term and d is common difference
$$\eqalign{ & d = {S_n} - k{S_{n - 1}} + {S_{n - 2}} \cr & \Rightarrow k{S_{n - 1}} = {S_n} + {S_{n - 2}} - d \cr & = \left( {{a_n} + {S_{n - 1}}} \right) + \left( {{S_{n - 1}} - {a_{n - 1}} - 1} \right) - d \cr} $$

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l} {∵ S_n} = {S_{n - 1}} + {a_n}\,{\rm{and}}\\ {S_{n - 1}} = {a_{n - 1}} + {S_{n - 2}}\\ \Rightarrow {S_{n - 2}} = {S_{n - 1}} - {a_{n - 1}} \end{array} \right\}\]

$$\eqalign{ & = {a_n} + 2{S_{n - 1}} - {a_{n - 1}} - d \cr & = 2{S_{n - 1}} + {a_n} - {a_{n - 1}} - d \cr & = 2{S_{n - 1}} + d - d\,\,\left( {\because {a_n} - {a_{n - 1}} = d} \right) \cr & = 2{S_{n - 1}} \cr & \therefore k = 2 \cr} $$