$${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$$

$$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}^2}} \right)$$

$$ = {{16} \over {25}}\left( {{{11\left( {11 + 1} \right)\left( {22 + 1} \right)} \over 6} - 1} \right)$$

$$ = {{16} \over {25}} \times 505 = {{16} \over 5} \times 101$$

$$ \Rightarrow {{16} \over 5}m = {{16} \over 5} \times 101$$

$$ \Rightarrow m = 101.$$

Let the $$GP$$ be $$a,ar$$ and $$a{r^2}$$ then

$$a=A+d;$$ $$ar=A+4d;$$ $$a{r^2} = A + 8d$$

$$ \Rightarrow {{a{r^2} - ar} \over {ar - a}} = {{\left( {A + 8d} \right) - \left( {A + 4d} \right)} \over {\left( {A + 4d} \right) - \left( {A + d} \right)}}$$

$$r = {4 \over 3}$$

$$m = {{l + n} \over 2}$$ and common ratio of

$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$

$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$

$$G_1^4 + 2G_2^4 + G_3^4$$

$$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$

$$ = \ln {\left( {1 + n} \right)^2}$$

$$ = \ln \times 2{m^2}$$

$$ = 4l{m^2}n$$

$${n^{th}}$$ term of series

$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$

Sum of $$n$$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$

$$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$$

$$ = {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$$

Sum of $$9$$ terms

$$ = {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$