Question

Find three consecutive terms in a G.P. such that the sum of the first two terms is $9$ and the product of all the three is $216$.

• A. $1,8,19$
• B. $2,7,13$
  
• C. $3,6,12$
• D. $4,5,16$
  

Answer 1

Answer: (C)

$3,6,12$

Let first three consecutive terms are in GP : $\frac{a}{r},a,ar$
Given, Sum of first two $=9$
$\therefore \frac{a}{r}+a=9$ ....... $\left(1\right)$
Product of first three terms $=216$
$\therefore \frac{a}{r}\times a\times ar=216$
$⟹a^3=216$
$⟹a^3=6^3$
$⟹a=6$

By putting $a=6$ in $\left(1\right),$ we get,

$\frac{6}{r}+6=9$
$⟹6(\frac{1}{r}+1)=9$
$⟹\frac{1}{r}+1=\frac{3}{2}$

$⟹\frac{1}{r}=\frac{3}{2}-1$

$⟹\frac{1}{r}=\frac{1}{2}$

$⟹r=2$
Now put the value of $a$ and $r$ in $\frac{a}{r},a,ar$,we get $3,6,\mathrm{12....}$
$\therefore$ Three consecutive terms of GP are $3,6,12$.