JEE PYQ: Sequence And Series Question 37
Question 37 - 2020 (07 Jan Shift 2)
Let $a_1, a_2, a_3, \ldots$ be a G.P. such that $a_1 < 0$, $a_1 + a_2 = 4$ and $a_3 + a_4 = 16$. If $\sum_{i=1}^{9} a_i = 4\lambda$, then $\lambda$ is equal to:
(1) $-513$
(2) $-171$
(3) 171
(4) $\frac{511}{3}$
Show Answer
Answer: (2)
Solution
Since, $a_1 + a_2 = 4 \Rightarrow a_1 + a_1r = 4$
$a_3 + a_4 = 16 \Rightarrow a_1r^2 + a_1r^3 = 16$
$r^2 = 4, a_1(1+r) = 4$
$r = -2, a_1(1-2) = 4 \Rightarrow a_1 = -4$
$\sum_{i=1}^{9} a_i = \frac{a(r^9 - 1)}{r - 1} = \frac{(-4)((-2)^9 - 1)}{-2 - 1}$
$= \frac{4}{3}(-513) = 4\lambda$
$\Rightarrow \lambda = -171$
BETA
