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

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

Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$

Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$

Multiplied by $${{11} \over {10}}$$ on both the sides

$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$

$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$

$$ \Rightarrow x = {10^{11}} = k{.10^9}$$

Given $$ \Rightarrow k = 100$$