Question

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 $\displaystyle\sum_{ k =1}^{ n } k ( n - k )$ is equal to :

• A. 560
• B. 1540
• C. 1330
• D. 2600

Answer 1

Answer: (C)

$ \left(\left(2^1 2^2 \ldots .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} $
$\displaystyle \sum_{ k =1}^{ n } nk - k ^2=\frac{ n ^2( n +1)}{2}-\frac{ n ( n +1)(2 n +1)}{6} $
$=1330$