Question

Let $S_n$ denote 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 $k=?$

Answer 1

Given: $d=S_n–k{S}_{n–1}+S_{n–2}$

Let $n=3$. Then, the AP becomes, $a,a+d,a+2d$. Now, for this AP,

$\Rightarrow S_n=S_3=\left(a\right)+(a+d)+(a+2d)=3a+3d=3(a+d)$

$\Rightarrow S_{n-1}=S_2=\left(a\right)+(a+d)=2a+d$

$\Rightarrow S_{n-2}=S_1=a$

$\therefore d=S_3–k{S}_2+S_1$

$\Rightarrow d=3(a+d)-k(2a+d)+a$

$\Rightarrow d=4a+3d-2ak-dk$

$\Rightarrow 0=4a+2d-2ak-dk$

$\Rightarrow 0=2a(2-k)+d(2-k)$

$\Rightarrow (2-k)(2a+d)=0$

If $(2-k)=0$, then,

$\Rightarrow 2-k=0$

$\Rightarrow k=2$

Therefore, value of $k$ is $2$.