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8. Sequences and Series
hard
Consider two G.Ps. $2,2^{2}, 2^{3}, \ldots$ and $4,4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{ k =1}^{ n } k (n- k )$ is equal to.
A
$560$
B
$1540$
C
$1330$
D
$2600$
(JEE MAIN-2022)
Solution
$\left(\left(2^{1} 2^{2} \cdots 2^{60}\right)\left(4^{1} \cdot 4^{2} \ldots \ldots 4^{ n }\right)\right)^{\frac{1}{60+ n }}=2^{\frac{225}{8}}$
$\left(2^{30 \times 61} 4^{\frac{ n ( n +1)}{2}}\right) \frac{1}{60+ n }=2^{\frac{225}{8}}$
$2^{1830+ n ^{2}+ n }=2^{\frac{(225)(60+ n )}{8}}$
$=8 n ^{2}-217 n +1140=0$
$n =20, \frac{57}{8}$
$\sum_{ k =1}^{ n } nk – k ^{2}=\frac{ n ^{2}( n +1)}{2}-\frac{ n ( n +1)(2 n +1)}{6}$
$=1330$
Standard 11
Mathematics
