Question

Find four numbers forming a geometric progression in which the sum of the extreme terms is $112$ and the sum of the middle terms is $48$.

Answer 1

Let the geometric progression be $a,ar,{ar}^2\&{ar}^3$

$\therefore a+a{r}^3=112\Rightarrow a(1+r^3)=112-\left(i\right)\&ar+a{r}^2=48\Rightarrow ar(1+r)=48-\left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$

$\frac{a(1+r^3)}{ar(1+r)}=\frac{112}{48}\Rightarrow \frac{a(1+r)(1+r+r^2)}{ar(1+r)}=\frac{112}{48}\Rightarrow \frac{1+r+r^2}{r}=\frac{7}{3}\Rightarrow 3+3r+3r^2=7r\Rightarrow 3r^2-4r+3=0a=3,b=-4,c=3\therefore r=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\times 3\times 3}}{2\times 3}=\frac{4\pm \sqrt{16-36}}{5}=\frac{4\pm \sqrt{-20}}{5}=\frac{4\pm 2\sqrt{5}i}{5}$