Let $S$ be the sum of the first 9 terms of the series:
$\{x+k a\}+\left\{x^{2}+(k+2) a\right\}+\left\{x^{3}+(k+4) a\right\}+$
$\left\{x^{4}+(k+6) a\right\}+\ldots . .$ where $a \neq 0$ and $x \neq 1 .$ If
$S=\frac{x^{10}-x+45 a(x-1)}{x-1}$, then $k$ is equal to :
Correct Option: , 3
$S=[x+k a+0]+\left[x^{2}+k a+2 a\right]+\left[x^{3}+k a+\right.$
$4 a]+\left[x^{4}+k a+6 a\right]+\ldots .9$ terms
$\Rightarrow S=\left(x+x^{2}+x^{3}+x^{4}+\ldots . .9\right.$ terms $)+(k a+k a$
$+k a+k a+\ldots \ldots .9$ terms $)+(0+2 a+4 a+6 a+$
9 terms)
$\Rightarrow \mathrm{S}=\mathrm{x}\left[\frac{\mathrm{x}^{9}-1}{\mathrm{x}-1}\right]+9 \mathrm{ka}+72 \mathrm{a}$
$\Rightarrow S=\frac{\left(x^{10}-x\right)+(9 k+72) a(x-1)}{(x-1)}$
Compare with given sum, then we get, $(9 \mathrm{k}+$ 72) $=45$
$\Rightarrow \mathrm{k}=-3$