Let S be the sum of the first 9 terms of the | vedclass

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(C) The given series is $S = (x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + \dots$ up to $9$ terms.We can split the sum into three parts:$S = (x + x^2 + x^3 +

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$-3$

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(C) The given series is $S = (x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + \dots$ up to $9$ terms.We can split the sum into three parts:$S = (x + x^2 + x^3 + \dots + x^9) + (ka + ka + \dots + ka 	ext{ [9 times]}) + (0 + 2a + 4a + \dots + 16a)$.The first part is a geometric progression: $x \frac{x^9-1}{x-1} = \frac{x^{10}-x}{x-1}$.The second part is $9ka$.The third part is an arithmetic progression with $n=9$,first term $0$,and common difference $2a$: $\frac{9}{2} [2(0) + (9-1)2a] = \frac{9}{2} [16a] = 72a$.Thus,$S = \frac{x^{10}-x}{x-1} + 9ka + 72a = \frac{x^{10}-x + (9k+72)a(x-1)}{x-1}$.Comparing this with the given expression $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,we get $9k + 72 = 45$.$9k = 45 - 72 = -27$.$k = -3$.

Let $S$ be the sum of the first $9$ terms of the series: $(x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + (x^4+(k+6)a) + \dots$ 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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