Question

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

• A. $4l^{2}mn$
• B. $4l{m}^{2}n$
• C. $lm{n}^2$
• D. $l^2{m}^2{n}^2$

Answer 1

Answer: (B)

Given, $m$ is the AM of $l$ and $n$.
$\therefore l+n=2m...(i)$
and $G_1,G_2,G_3$ are geometric means between $l$ and $n$.
$l,G_1,G_2,G_3$ are in GP.
Let $r$ be the common ratio of this GP.
$\therefore G_1=lr,G_2=l{r}^2,G_3=l{r}^3,n-l{r}^4\Rightarrow r=(\frac{n}{l}{)}^{\frac{1}{4}}$
Now,$G_1^4+2G_2^4+G_3^4=(lr{)}^4+2(l{r}^2{)}^4+(l{r}^3{)}^4$
$=l^4\times r^4(1+2r^4+r^6)=l^4\times r^4+(r^4+1{)}^2$
$=l^4\times \frac{n}{l}(\frac{n+l}{l}{)}^2=ln\times 4m^2=4l{m}^{2}n$