Answer & Solution
Answer: Option B
Solution:
S
n is the sum of n terms of an A.P.
a is its first term and d is common difference
$$\eqalign{
& d = {S_n} - k{S_{n - 1}} + {S_{n - 2}} \cr
& \Rightarrow k{S_{n - 1}} = {S_n} + {S_{n - 2}} - d \cr
& = \left( {{a_n} + {S_{n - 1}}} \right) + \left( {{S_{n - 1}} - {a_{n - 1}} - 1} \right) - d \cr} $$
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}
{∵ S_n} = {S_{n - 1}} + {a_n}\,{\rm{and}}\\
{S_{n - 1}} = {a_{n - 1}} + {S_{n - 2}}\\
\Rightarrow {S_{n - 2}} = {S_{n - 1}} - {a_{n - 1}}
\end{array} \right\}\]
$$\eqalign{
& = {a_n} + 2{S_{n - 1}} - {a_{n - 1}} - d \cr
& = 2{S_{n - 1}} + {a_n} - {a_{n - 1}} - d \cr
& = 2{S_{n - 1}} + d - d\,\,\left( {\because {a_n} - {a_{n - 1}} = d} \right) \cr
& = 2{S_{n - 1}} \cr
& \therefore k = 2 \cr} $$
