Question

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

The Correct Option is 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.
$\log a -\log 2 b , \log 2 b -\log 3 c , \log 3 c -\log a$ are in A.P.
$\Rightarrow 2(\log 2 b -\log 3 c )=(\log a -\log 2 b )+(\log 3 c -\log a ) $
$\Rightarrow 3 \log 2 b =3 \log 3 c \Rightarrow 2 b =3 c$
Now, $b^{2}=a c$
$ \Rightarrow b^{2}=a \cdot \frac{2 b}{3} $
$\Rightarrow b=\frac{2 a}{3}, c=\frac{4 a}{9} $
i.e., $ a=a, b=\frac{2 a}{3}, c=\frac{4 a}{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.