JEE PYQ: Sequence And Series Question 32
Question 32 - 2020 (05 Sep Shift 2)
If the sum of the first 20 terms of the series $\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + \ldots$ is 460, then $x$ is equal to:
(1) $7^2$
(2) $7^{1/2}$
(3) $e^2$
(4) $7^{46/21}$
Show Answer
Answer: (1)
Solution
$S = \log_7 x^2 + \log_7 x^3 + \log_7 x^4 + \ldots$ 20 terms
$\therefore S = 460$
$\Rightarrow \log_7(x^2 \cdot x^3 \cdot x^4 \cdot \ldots \cdot x^{21}) = 460$
$\Rightarrow (2 + 3 + 4 + \ldots + 21)\log_7 x = 460$
$\Rightarrow \frac{20}{2}(2 + 21)\log_7 x = 460$
$\Rightarrow \log_7 x = \frac{460}{230} = 2 \Rightarrow x = 7^2 = 49$
BETA
