Question

Consider two G.Ps. $2,2^2,2^3,\dots$ and $4,4^2$, $4^3,\dots$ 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

Answer 1

Answer: (C)

${\left(\left(2^1{2}^2\dots .2^{60}\right)\left(4^1\cdot 4^2\dots \dots 4^n\right)\right)}^{\frac{1}{60+n}}=2^{\frac{225}{8}}$
$\left(2^{30\times 61}{4}^{\frac{n(n+1)}{2}}\right)\frac{1}{60+n}=2^{\frac{225}{8}}$
$2^{1830+n^2+n}=2^{\frac{(225)(60+n)}{8}}$
$=8n^2-217n+1140=0$
$n=20,\frac{57}{8}$
$\sum _{k=1}^{n}nk-k^2=\frac{n^2(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$
$=1330$