Question

Let $S$ be the sum of the first $9$ terms of the series:

$\left\{x+ka\right\}+\left\{x^2+(k+2)a+x^3+(k+4)a+x^4+(k+6)a\right\}+\dots$ where $a\ne 0$ and $a\ne 1$.

If $S=\frac{(x^{10}-x+45a(x-1\left)\right)}{(x-1)}$, then $k$ is equal to:

• A. $3$
• B. $-3$
• C. $1$
• D. $5$

Answer 1

The correct option is B

$-3$

Explanation for the correct option:

Finding the value of $k$

The given series,

$\left\{x+ka\right\}+\left\{x^2+(k+2)a+x^3+(k+4)a+x^4+(k+6)a\right\}+\dots$ where, $a\ne 0$ and $a\ne 1$

$$\begin{gathered} \Rightarrow \left\{x+ka\right\}+\left\{x^2+(k+2)a+x^3+(k+4)a+x^4+(k+6)a\right\}+\dots \\ \Rightarrow S=(x+x^2+\dots .+x^9)+a\left\{k+(k+2)+(k+4)+\dots \dots \right\}...\left(i\right) \end{gathered}$$ up to nine terms

We can see that it involved two series, G.P with first element $a=x$and common ratio $r=x$ and A.P. with first element $a=k$ common difference $d=2$

Since, sum of $n$ term of G. P $=\frac{a\left(1-r^n\right)}{1-r}$ where $a$ is the first element.

And sum of n term of A. P. = $\frac{n}{2}[2a+\left(n-1\right)d]$ where $a$ is first element and $l$ is last element

Therefore from (i),

$$\begin{gathered} \Rightarrow s=\frac{x\left(1-x^9\right)}{1-x}+a\left[\frac{9}{2}(2k+(9-1)\times 2\right] \\ \Rightarrow s=\frac{x\left(1-x^9\right)}{1-x}+a\left[9\left(k+8\right)\right] \\ \Rightarrow s=\frac{x\left(1-x^9\right)+a\left(9k+72\right)\left(1-x\right)}{-(x-1)} \end{gathered}$$

Since,

$S=\frac{(x^{10}-x+45a(x-1\left)\right)}{(x-1)}$ is given

$$\begin{gathered} \Rightarrow \frac{(x^{10}-x+45a(x-1\left)\right)}{(x-1)}=\frac{x\left(1-x^9\right)+a\left(9k+72\right)\left(1-x\right)}{-\left(x-1\right)} \\ \Rightarrow -(x^{10}-x+45a(x-1\left)\right)=x-x^{10}+a\left(9k+72\right)\left(1-x\right) \\ \Rightarrow 45=9k+72 \\ \Rightarrow 9k=-27 \\ \Rightarrow k=-3 \end{gathered}$$

Hence, the correct option is (B)