JEE PYQ: Sequence And Series Question 14
Question 14 - 2021 (26 Feb Shift 1)
In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25. Then, the sum of $4^{th}$, $6^{th}$ and $8^{th}$ terms is equal to:
(1) 35
(2) 30
(3) 26
(4) 32
Show Answer
Answer: (1)
Solution
$ar + ar^5 = \frac{25}{2}$
$ar^2 \times ar^4 = 25$
$a^2r^6 = 25$
$a = \frac{5}{r^3}$
$\frac{5r}{r^3} + \frac{5r^5}{r^3} = \frac{25}{2}$
$\frac{1}{r^2} + r^2 = \frac{5}{2}$
Put $r^2 = t$
$\frac{t^2+1}{t} = \frac{5}{2}$
$2t^2 - 5t + 2 = 0$
$(2t - 1)(t - 2) = 0$
$t = \frac{1}{2}, 2 \Rightarrow r^2 = \frac{1}{2}, 2$
$r = \sqrt{2}$
$= ar^3 + ar^5 + ar^7$
$= ar^3(1 + r^2 + r^4)$
$= 5[1 + 2 + 4] = 35$
BETA
