If m is the A.M.

Question

If $m$ is the A.M. of two distinct real numbers $l$ and $n\ (l,n>1)$, and $G_1,G_2,G_3$ are three geometric means between $l$ and $n$, then $G_1^4+2G_2^4+G_3^4$ equals:

Accepted Answer

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

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

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

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

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

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

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

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

$= 4l{m^2}n$

Solutions · 1