JEE MAIN - Mathematics (2015 (Offline) - No. 17)
If m is the A.M. of two distinct real numbers l and n $$(l,n > 1)$$ and $${G_1},{G_2}$$ and $${G_3}$$ are three geometric means between $$l$$ and n, then $$G_1^4\, + 2G_2^4\, + G_3^4$$ equals:
$$4\,lm{n^2}$$
$$4\,{l^2}{m^2}{n^2}$$
$$4\,{l^2}m\,n$$
$$4\,l\,{m^2}n$$
Explanation
$$m = {{l + n} \over 2}$$ and common ratio of
$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$
$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$
$$G_1^4 + 2G_2^4 + G_3^4$$
$$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$
$$ = \ln {\left( {1 + n} \right)^2}$$
$$ = \ln \times 2{m^2}$$
$$ = 4l{m^2}n$$
$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$
$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$
$$G_1^4 + 2G_2^4 + G_3^4$$
$$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$
$$ = \ln {\left( {1 + n} \right)^2}$$
$$ = \ln \times 2{m^2}$$
$$ = 4l{m^2}n$$
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