Question

If $\log a,\log b$, and $\log c$ are in A.P. and also $\log a-\log 2b,\log 2b-\log 3c,\log 3c-\log a$ are in A.P., then

• A. a, b, c, are in H.P.
• B. a, 2b, 3c are in A.P.
• C. a, b, c are the sides of a triangle
• D. none of the above

Answer 1

Answer: (C)

$\log a,\log b,\log c$ are in A.P.
$\Rightarrow 2\log b=\log a+\log c$
$\Rightarrow \log b^2=\log (ac)$
$\Rightarrow b^2=ac\Rightarrow a,b,c$ are in G.P.
$\log a-\log 2b,\log 2b-\log 3c,\log 3c-\log a$ are in A.P.
$\Rightarrow 2(\log 2b-\log 3c)=(\log a-\log 2b)+(\log 3c-\log a)$
$\Rightarrow 3\log 2b=3\log 3c\Rightarrow 2b=3c$
Now, $b^2=ac$
$\Rightarrow b^2=a\cdot \frac{2b}{3}$
$\Rightarrow b=\frac{2a}{3},c=\frac{4a}{9}$
i.e., $a=a,b=\frac{2a}{3},c=\frac{4a}{9}$
$\Rightarrow a:b:c=1:\frac{2}{3}:\frac{4}{9}=9:6:4$
Since, sum of any two is greater than the $3^{rd},a,b,c$ form a triangle.