Question

If $S_1,S_2,S_3,...,S_n$ denote the sum to $1,2,3,...,n$ terms of an A.P. having first term $a$ and $\frac{S_{kx}}{S_x}$ is independent of $x,$ then prove $S_1+S_2+S_3+...+S_n=\frac{n(n+1)(2n+1)a}{6}$

Answer 1

$\frac{S_{kx}}{S_x}=\frac{\frac{kx}{2}[2a+(kx-1)d]}{\frac{x}{2}[2a+(x-1)d]}=\frac{k[2a-d+kxd]}{2a-d+xd}$

For this to be independent of $x,2a-d=0\Rightarrow 2a=d$

Now, $S_n=\frac{n}{2}[2a+(n-1)d]=\left(\frac{n^2}{2}\right).2a=a{n}^2$

$\therefore S_1+S_2+S_3+...+S_n=\frac{n(n+1)(2n+1)a}{6}$