Question

If $loga,logb$, and $logc$ are in A.P. and also $loga-log2b,log2b-log3c,log3c-loga$ 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 these

Answer 1

Answer: (C)

$loga,logb,logc$ are in A.P.
$\Rightarrow 2logb=loga+logc$
$\Rightarrow log{b}^2=log(ac)$
$\Rightarrow b^2=ac$
$\Rightarrow a,b,c$, are in G.P.
$loga-log2b,log2b-log3c,log3c-log$ 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.c. $a=a,b=\frac{2a}{3},c=\frac{4a}{9}$
$\Rightarrow a:b:c=1:\frac{2}{3}:\frac{4}{9}=9:6:4$
Since, sum of any two is greater than the $3rd,a,b,c$, form a triangle