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 means 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

The correct option is B $4l{m}^{2}n$

Given, m is the AM of l and n
$\therefore l+n=2m\dots \dots \left(i\right)$
and $G_1,G_2,G_3$ are geometric means between l and n
$l,G_1,G_2,G_3,n$ are in GP
Let r be the common ratio of this GP
$\therefore G_1=lr,G_2=l{r}^2,G_3=l{r}^3,n=l{r}^4$
$\Rightarrow r={\left(\frac{n}{l}\right)}^{\frac{1}{4}}$
Now, $G_1^4+2G_2^4+G_3^4=(lr{)}^4+2(l{r}^2{)}^4+(l{r}^3{)}^4$
$=l^4\times r^4(1+2r^4+r^8)=l^4\times r^4(r^4+1{)}^2$
$=l^4\times \frac{n}{l}{\left(\frac{n+l}{l}\right)}^2=ln\times 4m^2=4l{m}^{2}n$