Question

The product of three numbers in $GP$ is $1000.$ If $6$ is added to the second number and $7$ is added to the third number, we get an $AP.$ Find the numbers ?

Answer 1

Solution:

Let the $3$ numbers be $\frac{a}{r},a,ar$.

$\frac{a}{r}\times a\times ar=1000$

$a^3=1000$

$a=10$

$2(a+6)=(ar+7)+\frac{a}{r}$

Substituting $a=10$, we get

$32r=10r^2+7r+10$

$10r^2-25r+10$

$2r^2-5r+2=0$

$(r-2)(2r-1)=0$

$r=2,\frac{1}{2}$

Numbers are $5,10,20$ or $20,10,5$

(r−2)(2r−1)=0r=2