Question

If $a,b,c$ are in geometric progression with common ratio $r(r>1)$ such that $abc=216$ and $ab+bc+ca=156,$ then

• A. $a+c=20,a<c$
• B. $a+c=20,a>c$
• C. $a+c=24,a>c$
• D. $a+c=24,a<c$

Answer 1

The correct option is A $a+c=20,a<c$

$a,b,c$ are in G.P.
Common ratio $=r(r>1)$
Let $a=\frac{b}{r}$ and $c=br$
$abc=216\Rightarrow b^3=216$
$\Rightarrow b=6$

$ab+bc+ca=156$
$\Rightarrow b^2(\frac{1}{r}+r+1)=156\Rightarrow 3r^2-10r+3=0$
$\Rightarrow r=\frac{1}{3}$ or $3$
But $r>1$
$\therefore r=3$
$\Rightarrow a=2$ and $c=18$