Question

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 :

• A. -5
• B. 1
• C. -3
• D. 3

Answer 1

Answer: (C)

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