Thank you for reporting, we will resolve it shortly
Q.
If $a, b, c$ are positive integers forming an increasing G.P. and $b-a$ is a perfect cube and $\log _6 a+\log _6 b+\log _6 c=6$, then $a+b+c$ is equal to
Sequences and Series
Solution:
$\Theta \log _6 a b c=6 \Rightarrow a b c=6^6 \Rightarrow b^3=6^6 \Rightarrow b=6^2=36$
If $r$ is common ratio then terms are $\frac{36}{ r }, 36,36 r$
$\Theta \frac{36}{r} \in I$
$\therefore $ r can be $2,3,4,6,9,12,18$
and $ 36-\frac{36}{ r }=36\left(1-\frac{1}{ r }\right)$ is perfect cube
$\therefore r=4$
$\therefore a+b+c=9+36+144=189$