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(C) The series is given by $S = \sum_{n=1}^{9} [x^n + (k + 2(n-1))a]$.Expanding the sum,we get $S = (x + x^2 + \ldots + x^9) + \sum_{n=1}^{9} (k + 2n

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$-3$

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(C) The series is given by $S = \sum_{n=1}^{9} [x^n + (k + 2(n-1))a]$.Expanding the sum,we get $S = (x + x^2 + \ldots + x^9) + \sum_{n=1}^{9} (k + 2n - 2)a$.The sum of the geometric progression is $\sum_{n=1}^{9} x^n = x \frac{x^9 - 1}{x - 1} = \frac{x^{10} - x}{x - 1}$.The sum of the arithmetic part is $\sum_{n=1}^{9} (k + 2n - 2)a = a [9k + 2 \frac{9 	imes 8}{2} - 18] = a [9k + 72 - 18] = a(9k + 54)$.Wait,recalculating the arithmetic part: $\sum_{n=0}^{8} (k + 2n)a = 9ka + 2a \frac{8 	imes 9}{2} = 9ka + 72a = a(9k + 72)$.Thus,$S = \frac{x^{10} - x}{x - 1} + \frac{a(9k + 72)(x - 1)}{x - 1} = \frac{x^{10} - x + (9k + 72)a(x - 1)}{x - 1}$.Comparing this with the given $S = \frac{x^{10} - x + 45a(x - 1)}{x - 1}$,we have $9k + 72 = 45$.$9k = 45 - 72 = -27$.$k = -3$.

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 $a \neq 0$ and $x \neq 1$. If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,then $k$ is equal to:

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