Let $S$ be the sum of the first 9 terms of the series : $\{x+k a\}+\left\{x^{2}+(k+2) a\right\}+\left\{x^{3}+(k+4) a\right\}$
$+\left\{x^{4}+(k+6) a\right\}+\ldots \quad$ where $\quad a \neq 0$ and $x \neq 1$. If $S=\frac{x^{10}-x+45 a(x-1)}{x-1}$, then $k$ is equal to :
Correct Option: , 3
$S=\left(x+x^{2}+x^{3}+\ldots .9\right.$ terms $)$
$+a[k+(k+2)++(k+4)+\ldots 9$ terms $]$
$\Rightarrow S=\frac{x\left(x^{9}-1\right)}{x-1}+\frac{9}{2}[2 a k+8 \times(2 a)]$
$\Rightarrow S=\frac{x^{10}-x}{x-1}+\frac{9 a(k+8)}{1}=\frac{x^{10}-x+45 a(x-1)}{x-1}$ (Given)
$\Rightarrow \frac{x^{10}-x+9 a(k+8)(x-1)}{x-1}=\frac{x^{10}-x+45 a(x-1)}{x-1}$
$\Rightarrow 9 a(k+8)=45 a \Rightarrow k+8=5 \Rightarrow k=-3$
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