Sequences & Series
GP with perfect cube and logarithm conditions
MJMT_Full_Test_10
Grade 12
Question:
If $a$, $b$, $c$ are positive integers forming an increasing GP, $b-a$ is a perfect cube, and $\log_6 a + \log_6 b + \log_6 c = 6$, then $a+b+c$ is equal to
Step-by-Step Solution
Key Concept: Let $a=b/r$, $c=br$ for common ratio $r$. Then $\log_6(abc)=6 \Rightarrow abc=6^6 \Rightarrow b^3=6^6 \Rightarrow b=36$.
$b=36$, $r=4$: $a=9$, $c=144$. $b-a=27=3^3$ ✓. $a+b+c=189$.
Correct Answer: 4