QJEEM22Ma7aa3: JEE Main 2022 Mathematics - Class 11 JEEM22Ma7aa3 Solution: Consider two G.Ps. 2, 2 2 , 2 3 , ..... and 4, 4 2 , 4 3 , .... …

Given G.P's 2, 2², 2³, .... 60 terms

4, 4², .... n terms

Now, G.M $= {2^{{{225} \over 8}}}$

${\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}$

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

$\Rightarrow {{{n^2} + n + 1830} \over {60 + n}} = {{225} \over 8}$

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

$n = {{57} \over 8},\,20,\,$ so $n = 20$

$\therefore$ $\sum\limits_{k = 1}^{20} {k(20 - k) = 20 \times {{20 \times 21} \over 2} - {{20 \times 21 \times 41} \over 6}}$

$= {{20 \times 21} \over 2}\left[ {20 - {{41} \over 3}} \right] = 1330$