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

Given, $m$m is the A.M. of l&n.

$$\Rightarrow m=\frac{l+n}{2}...(1)$$⇒m=2l+n​...(1)

${G}_{1},{G}_{2},{G}_{3}$G1​,G2​,G3​ are G.M. between l&n.

$\Rightarrow l,{G}_{1},{G}_{2},{G}_{3},n$⇒l,G1​,G2​,G3​,n are in G.P.

$$\Rightarrow n=l{(r)}^{4}\Rightarrow {r}^{4}=\frac{n}{l}...(2)$$⇒n=l(r)4⇒r4=ln​...(2)

$$\Rightarrow {G}_{1}=l⁡r,{G}_{2}=l⁡{r}^{2},{G}_{3}=l⁡{r}^{3}$$⇒G1​=l⁡r,G2​=l⁡r2,G3​=l⁡r3

$$\Rightarrow {G}_{1}^{4}+2{G}_{2}^{4}+{G}_{3}^{4}={l⁡}^{4}{r}^{4}+2\times {l⁡}^{4}{r}^{8}+{l⁡}^{4}\times {r}^{12}$$⇒G14​+2G24​+G34​=l⁡4r4+2×l⁡4r8+l⁡4×r12

$$={l⁡}^{4}{r}^{4}[1+2{r}^{4}+{r}^{8}]$$=l⁡4r4[1+2r4+r8]

$$={l}^{4}\times \frac{n}{l⁡}{[1+{r}^{4}]}^{2}={l⁡}^{3}n\times {[1+\frac{n}{l⁡}]}^{2}$$=l4×l⁡n​[1+r4]2=l⁡3n×[1+l⁡n​]2

$$={l⁡}^{3}\times n\times \frac{{(l⁡+n)}^{2}}{{l⁡}^{2}}$$=l⁡3×n×l⁡2(l⁡+n)2​

$=l⁡n\times {(2m)}^{2}$=l⁡n×(2m)2 (From (1))

$$=4l⁡{m}^{2}n$$=4l⁡m2n