Question

Three numbers $a,b$ and $12$ form in that order, a geometric progression and the numbers $a,b$ and $9$ form an arithmetic progression. Find $a$ and $b$.

Answer 1

a,b & 12 are in geometric progression.

$\therefore a_1=a,a_{1}r=b$ where $a_1=1^{st}term,r=commonratio$

$\Rightarrow r=\frac{b}{a}$

$\frac{12}{b}=r$

$\Rightarrow \frac{12}{b}=\frac{b}{a}$

$\Rightarrow b^2=12a$-----------------1

Also, a,b & 9 form an AP.

$\therefore b-a=9-b$

$\Rightarrow 2b=9+a$

$\Rightarrow a=2b-9$---------------------------------2

From Equation 1 & Equation 2, we get

$b^2=12(2b-9)$

$\Rightarrow b^2=24b-108$

$\Rightarrow b^2-24b+108=0$

$\Rightarrow b^2-18b-6b+108=0$

$\Rightarrow b(b-18)-6(b-18)=0$

$\Rightarrow b=6$ or $b=18$.

When $b=6$,

$a=\frac{b^2}{12}=\frac{6\times 6}{12}=3$

When $b=18$,

$a=\frac{b^2}{12}=\frac{18\times 18}{12}=27$