JEE PYQ: Sequence And Series Question 68
Question 68 - 2019 (12 Jan Shift 1)
If the sum of the first 15 terms of the series $\left(\frac{3}{4}\right)^3 + \left(1\frac{1}{2}\right)^3 + \left(2\frac{1}{4}\right)^3 + 3^3 + \left(3\frac{3}{4}\right)^3 + \ldots$ is equal to $225k$ then $k$ is equal to:
(1) 108
(2) 27
(3) 54
(4) 9
Show Answer
Answer: (2)
Solution
$S = \left(\frac{3}{4}\right)^3 + \left(\frac{3}{2}\right)^3 + \left(\frac{9}{4}\right)^3 + (3)^3 + \left(\frac{15}{4}\right)^3 + \ldots$
Let the general term of $S$ be $T_r = \left(\frac{3r}{4}\right)^3$, then
$255K = \sum_{r=1}^{15} T_r = \left(\frac{3}{4}\right)^3 \sum_{r=1}^{15} r^3$
$255K = \frac{27}{64} \times \left(\frac{15 \times 16}{2}\right)^2$
$\Rightarrow K = 27$
BETA
