Question

If $m$ is the arithmetic mean 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. $4lm{n}^2$
• B. $4l^2{m}^2{n}^2$
• C. $4l^{2}mn$
• D. $4l{m}^{2}n$

Answer 1

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

Given $G_1,G_2,G_3$ are three geometric means between $l$ and $n.$
$\Rightarrow l,G_1,G_2,G_3,n$ are in G.P.
Let $r$ be the common ratio.
Then $G_1=lr,G_2=l{r}^2,G_3=l{r}^3,n=l{r}^4$

Also, $m$ is the arithmetic mean between $l$ and $n.$
$\Rightarrow m=\frac{l+n}{2}$
$\Rightarrow 2m=l+l{r}^4\cdots \left(1\right)$

Hence, $G_1^4+2G_2^4+G_3^4$
$=l^4{r}^4+2l^4{r}^8+l^4{r}^{12}$
$=l^4{r}^4(1+2r^4+r^8)$
$=l^4{r}^4(1+r^4{)}^2$
$=l^4{r}^4{\left(\frac{2m}{l}\right)}^2\left[\text{From equation}\right(1\left)\right]$
$=n\cdot l^3\frac{4m^2}{l^2}$
$=4l{m}^{2}n$