Question

Find three numbers in G.P. whose sum is 65 and whose product is 3375.

Answer 1

Let the three number in G.P. be $\frac{a}{r}$, a, ar sum of these numbers $=\frac{a}{r}+a+ar=65$

3375 = product of the G.P.

$3375=\left(\frac{a}{r}\right(a\left)\right(ar)=a^3)$

$a^3=(5{)}^3\times (3{)}^3=(15{)}^3$

$\Rightarrow a=15$

$\Rightarrow a(\frac{1}{r}+1+r)=65$

Substituting the value of a

$\Rightarrow 15(\frac{1}{r}+1+r)=65$

$\Rightarrow \frac{1+r+r^2}{r}=\frac{13}{3}$

$\Rightarrow 3+3r+3r^2=13r$

$\Rightarrow 3r^2-10r+3=0$

$\Rightarrow 3r^2-r-9r+3=0$

$\Rightarrow r(3r-1)-3(3r-1)=0$

$r=\frac{1}{3}orr=3$

Hence, the G.P. for a = 15 and $r=\frac{1}{3}$ is 45, 15, 5

And the G.P. for a = 15 and r = 3 is 5, 15, 45