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  4. 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

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Q. 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

Sequences and Series

Solution:

$\Theta \log _6 a b c=6 \Rightarrow a b c=6^6$
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.
$\therefore a + b + c =9+36+144=189$.