Consider two G.Ps. 2, 2 ^ 2 , 2 ^ 3 , and 4, 4 ^ 2 , 4 ^ 3 , of 60...

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

Also given the geometric mean of all the $60 + n$ terms is \left(2\right)\dfrac{225}{8}$,</p><p>So, \left(\left({2}^{1}{2}^{2} \ldots . {2}^{60}\right)\left({4}^{1} \cdot {4}^{2} \ldots \ldots . {4}^{n}\right)\right)\dfrac{1}{60 + n} = 2\dfrac{225}{8}$

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

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

On comparing both side we get,

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

$⇒ 8{n}^{2} - 217n + 1140 = 0$

$⇒ n = 20, \dfrac{57}{8}$

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

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