Question

Find three consecutive terms in a G.P. such that the sum of the $2^{nd}$ and $3^{rd}$ terms is $60$ and the product of all the three is $8000$.

• A. $10,20,40$
• B. $30,40.50$
• C. $60,40.50$
• D. $25,35,45$

Answer 1

The correct option is A $10,20,40$

Let, three consecutive numbers in GP are: $\frac{a}{r},a,ar$
sum of $2^{nd}$ and $3^{rd}$ is $60$
$\Rightarrow a+ar=60$ .......(1)
product of all the three is $8000$
$\Rightarrow \frac{a}{r}\times a\times ar=8000$
$\Rightarrow a^3=8000$
$\Rightarrow a=20$
By putting $a=20$ in (1). we get
$20+20\times r=60$
$\Rightarrow 20\times r=40$
$\Rightarrow r=2$
terms are in GP are: $\frac{a}{r},a,ar=10,20,40$