Question

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.

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

Answer 1

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

It is given that, $m$ is the $A.M.$ of $l$ and $n.$

$\therefore$ $l+n=2m$ ------ ( 1 )

$G_1,G_2,G_3$ are geometric means between $l$ and $n$

$\therefore$ $l,G_1,G_2,G_3,n$ are in $GP.$

Let $r$ be the common ratio of this $G.P.$

$\therefore$ $G_1=lr$

$\Rightarrow$ $G_2=l{r}^2$

$\Rightarrow$ $G_3=l{r}^3$

$\Rightarrow$ $n=l{r}^4$

$\Rightarrow$ $r={\left(\frac{n}{l}\right)}^{\frac{1}{4}}$ ---- ( 2 )

Now,

$\Rightarrow$ $G_1^4+2G_2^4+G_3^4=(lr{)}^4+2(l{r}^2{)}^4+(l{r}^3{)}^4$

$=l^4\times r^4(1+2r^4+r^8)$

$=l^4\times r^4(r^4+1{)}^2$

$=l^4\times \frac{n}{l}{\left(\frac{n+l}{l}\right)}^2$ [ From ( 2 ) ]

$=l^4\times \frac{n}{l}{\left(\frac{2m}{l}\right)}^2$ [ From ( 1 ) ]

$=ln\times 4m^2$

$=4l{m}^{2}n$