Consider two G.Ps. 2 , 2 2 , 2 3 , … and 4 , 4 2 , 4 3 , … of 60 and n terms respectively. If…

Given two G.Ps. $2,2^2,2^3,\dots$ and $4,4^2,4^3,\dots$ of $60$ and $n$ terms respectively,

Also given the geometric mean of all the $60+n$ terms is ${\left(2\right)}^{\frac{225}{8}}$,

So, ${\left(\left(2^1{2}^2\dots .2^{60}\right)\left(4^1·4^2\dots \dots .4^{\mathrm{n}}\right)\right)}^{\frac{1}{60+n}}=2^{\frac{225}{8}}$

$\Rightarrow {\left(2^{30\times 61}{4}^{\frac{n\left(n+1\right)}{2}}\right)}^{\frac{1}{60+n}}=2^{\frac{225}{8}}$

$\Rightarrow 2^{1830+n^2+n}=2^{\frac{\left(225\right)\left(60+n\right)}{8}}$

On comparing both side we get,

$\Rightarrow 1830+n^2+n=\frac{225\left(60+n\right)}{8}$

$\Rightarrow 8n^2-217n+1140=0$

$\Rightarrow n=20,\frac{57}{8}$

Now $\sum _{k=1}^{n}k\left(n-k\right)=$$\sum _{k=1}^{20}nk-k^2=\frac{n^2\left(n+1\right)}{2}-\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$=\frac{{20}^2\times 21}{2}-\frac{20\times 21\times 41}{6}=1330$