JEE PYQ: Sequence And Series Question 63
Question 63 - 2019 (11 Jan Shift 1)
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is:
(1) $\frac{1}{3}$
(2) $\frac{2}{3}$
(3) $\frac{2}{9}$
(4) $\frac{4}{9}$
Show Answer
Answer: (2)
Solution
Let the terms of infinite series are $a, ar, ar^2, ar^3, \ldots$
So, $\frac{a}{1-r} = 3$
Since, sum of cubes of its terms is $\frac{27}{19}$, that is sum of $a^3$, $a^3r^3, \ldots, \infty$ is $\frac{27}{19}$.
So, $\frac{a^3}{1-r^3} = \frac{27}{19}$
$\Rightarrow \frac{a}{1-r} \times \frac{a^2}{1+r^2+r} = \frac{27}{19}$
$\Rightarrow \frac{9(1-r)^2 \times 3}{1+r^2+r} = \frac{27}{19}$
$\Rightarrow 6r^2 - 13r + 6 = 0$
$\Rightarrow (3r-2)(2r-3) = 0$
$r = \frac{2}{3}, \frac{3}{2}$
As $|r| < 1$
So, $r = \frac{2}{3}$
BETA
