Question

The sum of four numbers in G.P. is $60$ and the A.M. between the first and the last is $18$. Find the numbers.

Answer 1

Four numbers in GP

$a,ar,a^2,a{r}^{3}a+ar+a{r}^2+a{r}^3=60\frac{a(r^4-1)}{r-1}=60\to \left(1\right)$

and

$\frac{a+a^3}{2}=18a(1+r^3)=36r=(\frac{36}{a}-1{)}^{\frac{1}{3}}$

The sum and AM are integer, so each term is +ve in GP For $a=1,2,3$ it is irrational

So $r=(\frac{36}{4}-1{)}^{\frac{1}{3}}=8^{\frac{1}{3}}=2$

Other term would make a rational and will not satisfy the sum

So, $a=4,r=2$

Numbers are

$a,ar,a{r}^2$ & $a{r}^3$

$=4,4.2,{4.2}^2$ & ${4.2}^3$

$=4,8,16,32$

Answer