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

We observe that every term of the given series is made up of two parts – a power of $$x$$ and an $$a$$-term whose coefficient increases by 2 each step. Writing the first few terms explicitly,

$$ \{x + ka\} + \{x^{2} + (k+2)a\} + \{x^{3} + (k+4)a\} + \{x^{4} + (k+6)a\} + \ldots $$

makes it clear that for the $$n^{\text{th}}$$ term (counting from 1) we have

$$ T_n \;=\; x^{\,n} \;+\; \bigl[k + 2(n-1)\bigr]\,a. $$

Because the question asks for the first nine terms, we must add $$T_1,T_2,\ldots,T_9$$.

Step 1 – Sum of the powers of $$x$$. For a geometric progression with common ratio $$x$$, the formula

$$ \sum_{r=1}^{N} x^{r} \;=\; \frac{x^{\,N+1}-x}{x-1}, \qquad (x \ne 1) $$

gives, for $$N = 9$$,

$$ \sum_{r=1}^{9} x^{r} \;=\; \frac{x^{10}-x}{x-1}. $$

Step 2 – Sum of the $$a$$-coefficients. The coefficient of $$a$$ in the $$n^{\text{th}}$$ term is $$k+2(n-1)$$. Hence

$$ \sum_{n=1}^{9}\bigl[k+2(n-1)\bigr] \;=\; \underbrace{\sum_{n=1}^{9} k}_{=\,9k} \;+\; \underbrace{\sum_{n=1}^{9} 2(n-1)}_{=\,2\sum_{m=0}^{8} m}. $$

We know that $$\sum_{m=0}^{8} m = \frac{8\cdot9}{2} = 36,$$ so

$$ \sum_{n=1}^{9}\bigl[k+2(n-1)\bigr] = 9k + 2 \times 36 = 9k + 72. $$

Therefore the sum of the $$a$$-parts equals $$a(9k+72).$$

Step 3 – Total sum $$S$$ of the first nine terms.

Combining the two pieces we obtained,

$$ S = \frac{x^{10}-x}{x-1} + a(9k+72). $$

To compare this with the expression supplied in the question we write the entire result over the common denominator $$x-1$$:

$$ S = \frac{x^{10}-x + a(9k+72)(x-1)}{x-1}. $$

Step 4 – Equating with the given expression. The question states that

$$ S = \frac{x^{10}-x + 45a(x-1)}{x-1}. $$

Since the numerators of two equal fractions with the same denominator must be identical, we set

$$ x^{10}-x + a(9k+72)(x-1) = x^{10}-x + 45a(x-1). $$

Subtracting $$x^{10}-x$$ from both sides gives

$$ a(9k+72)(x-1) = 45a(x-1). $$

Because $$a \ne 0$$ and $$x-1 \ne 0$$, we can cancel these factors, leaving the simple linear relation

$$ 9k + 72 = 45. $$

Solving for $$k$$,

$$ 9k = 45 - 72 = -27 \;\;\Longrightarrow\;\; k = \frac{-27}{9} = -3. $$

Hence, the correct answer is Option C.