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Q.
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 :
Sequences and Series
Solution:
$log\, a , log \,b , log \, c$ 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.
$log \,a-log \,2 b,log \,2 b-log 3c, log \,3c-log$ 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 2b=3 c$
Now, $b ^{2}= ac$
$ \Rightarrow b ^{2}= a \cdot \frac{2 b }{3} $
$\Rightarrow b =\frac{2 a }{3}$,
$c =\frac{4 a }{9}$
i.c. $a = a , b =\frac{2 a }{3}, c =\frac{4 a }{9}$
$ \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