Consider two G.Ps. 2, 2², 2³, ..... and 4, 4², 4³, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)^{{{225} \over 8}}}$, then $\sum\limits_{k = 1}^n {k(n - k)}$ is equal to :

<p>Given G.P's 2, 2², 2³, .... 60 terms</p> <p>4, 4², .... n terms</p> <p>Now, G.M $= {2^{{{225} \over 8}}}$</p> <p>$${\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}$$</p> <p>$$\left( {{2^{{{{n^2} + n + 1830} \over {60 + n}}}}} \right) = {2^{{{225} \over 8}}}$$</p> <p>$\Rightarrow {{{n^2} + n + 1830} \over {60 + n}} = {{225} \over 8}$</p> <p>$\Rightarrow 8{n^2} - 217n + 1140 = 0$</p> <p>$n = {{57} \over 8},\,20,\,$ so $n = 20$</p> <p>$\therefore$ $$\sum\limits_{k = 1}^{20} {k(20 - k) = 20 \times {{20 \times 21} \over 2} - {{20 \times 21 \times 41} \over 6}} $$</p> <p>$= {{20 \times 21} \over 2}\left[ {20 - {{41} \over 3}} \right] = 1330$</p>