If the sum of four numbers in GP is 60 and the arithmetic mean of the first and last numbers is 18 then the numbers are

Let four terms in a GP be $a r^3, a r, \frac{a}{r}$ and $\frac{a}{r^3}$. According to the given condition, $a r^3+a r+\frac{a}{r}+\frac{a}{r^3}=60$ ....(i) and $\quad \frac{a r^3+\frac{a}{r^3}}{2}=18$ $\Rightarrow \quad a r^3+\frac{a}{r^3}=36$ .....(ii) Now, from Eq. (i), we have $\left(a r+\frac{a}{r}\right)+a r^3+\frac{a}{r^3}=60$ $\Rightarrow \quad a\left(r+\frac{1}{r}\right)+36=60 \quad$ [from Eq. (ii)] $\Rightarrow \quad a\left(r+\frac{1}{r}\right)=24$ .....(iii) On dividing Eq. (iii) by Eq. (ii), we get $\frac{a\left(r^3+\frac{1}{r^3}\right)}{a\left(r+\frac{1}{r}\right)}=\frac{36}{24}$ $\Rightarrow \frac{\left(r+\frac{1}{r}\right)\left(r^2+\frac{1}{r^2}-1\right)}{r+\frac{1}{r}}=\frac{3}{2}$ $\Rightarrow \quad 2\left(r^2+\frac{1}{r^2}-1\right)=3$ $\Rightarrow \quad \frac{2\left(r^4+1-r^2\right)}{r^2}=3$ $\begin{array}{lc}\Rightarrow & 2 r^4+2-2 r^2=3 r^2 \\ \Rightarrow & 2 r^4-5 r^2+2=0 \\ \Rightarrow & 2 r^4-4 r^2-r^2+2=0 \\ \Rightarrow & 2 r^2\left(r^2-2\right)-1\left(r^2-2\right)=0 \\ \Rightarrow & \left(r^2-2\right)\left(2 r^2-1\right)=0 \\ \Rightarrow & r^2=2,2 r^2=1 \\ \Rightarrow & r= \pm \sqrt{2}, r= \pm \frac{1}{\sqrt{2}}\end{array}$ On putting $r=\sqrt{2}$ in Eq. (iii), we get $a\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)=24$ $\Rightarrow \quad a\left(\frac{2+1}{\sqrt{2}}\right)=24$ $\begin{aligned} \Rightarrow & 3 a & =24 \sqrt{2} \\ \Rightarrow & a & =8 \sqrt{2}\end{aligned}$ $\therefore$ Series becomes $8 \sqrt{2}(\sqrt{2})^3, 8 \sqrt{2}(\sqrt{2}), \frac{8 \sqrt{2}}{\sqrt{2}}$ and $\frac{8 \sqrt{2}}{(\sqrt{2})^3}$ i.e., $32,16,8$ and 4. If we take $r=\frac{1}{\sqrt{2}}$, we get the series 4, 8, 16 and 32.