JEE PYQ: Sequence And Series Question 21
Question 21 - 2020 (02 Sep Shift 2)
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:
(1) $-5$
(2) 1
(3) $-3$
(4) 3
Show Answer
Answer: (3)
Solution
$\Rightarrow S = \frac{x(x^9 - 1)}{x - 1} + \frac{9}{2}[2ak + 8 \times (2a)]$
$\Rightarrow S = \frac{x^{10} - x + 9a(k+8)(x-1)}{x-1}$
$\Rightarrow 9a(k + 8) = 45a \Rightarrow k + 8 = 5 \Rightarrow k = -3$.
BETA
