Question

If $a,b,c$ are positive integers forming an increasing G.P. and $b-a$ is perfect cube of an integer and ${\log }_{6}a+{\log }_{6}b+{\log }_{6}c=6$, then $a+b+c$ is equal to

• A. 100
• B. 111
• C. 122
• D. 189

Answer 1

Answer: (D)

$\Theta {\log }_{6}abc=6\Rightarrow abc=6^6$
Let G.P. be $\frac{b}{r}$, b, br
$\therefore b^3=6^6\Rightarrow b=6^2=36$
$\therefore \frac{36}{r}$ will be integer for $r=2,3,4,6,9,12,18$
and $b-a=36\left(1-\frac{1}{r}\right)$ is perfect cube for $r=4$ only.
$\therefore a+b+c=9+36+144=189$.