Question

Let $S$ be the sum of the first $9$ terms of the series: $\{x+ka\}+\left\{x^2+(k+2)a\right\}+\left\{x^3+(k+4)a\right\}+$ $\left\{x^4+(k+6)a\right\}+\dots ..$ where $a\ne 0$ and $x\ne 1.$ If $S=\frac{x^{10}-x+45a(x-1)}{x-1},$ then $k$ is equal to :

• A. -5
• B. 1
• C. -3
• D. 3

Answer 1

Answer: (C)

$S=[x+ka+0]+\left[x^2+ka+2a\right]+[x^3+ka+$
$+4a]+\left[x^4+ka+6a\right]+\dots 9$ terms
$\Rightarrow S=(x+x^2+x^3+x^4+\dots ..9$ terms $)$
$+(ka+kan+ka+ka+\dots \dots .9$ terms $)+(0+2a+4a+6a+\dots 9$ terms
$\Rightarrow S=x\left[\frac{x^9-1}{x-1}\right]+9ka+72a$
$\Rightarrow S=\frac{\left(x^{10}-x\right)+(9k+72)a(x-1)}{(x-1)}$
Compare with given sum, then we get, $(9k+72)=45$
$\Rightarrow k=-3$