Question

Find four numbers in $G.P.$ such that their products is $1$ and the sum of the middle two numbers is $\frac{10}{3}$.

Answer 1

Let the terms be $a,ar,a{r}^2,a{r}^3$

Product is $1⟹a\left(ar\right)\left(a{r}^2\right)\left(a{r}^3\right)=1$

$⟹a^4{r}^6=1⟹a=r^{-\frac{3}{2}}$

Sum of middle two terms is $\frac{10}{3}⟹ar+a{r}^2=\frac{10}{3}$

$⟹ar(1+r)=\frac{10}{3}⟹1+r=\frac{10}{3}\sqrt{r}$

On solving ,we get $r=9,\frac{1}{9}⟹a=\frac{1}{27},27$

the four numbers in $G.P.$ are $\frac{1}{27},\frac{1}{3},3,27$