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Question

If loga,logb, and logc are in A.P. and also logalog2b,log2blog3c,log3cloga are in A.P., then

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

The correct option is A a,b,c are the sides of a triangle
loga,logb,logc are in A.P.
2logb=loga+logc
logb2=log(ac)
b2=aca,b,c are in G.P.
logalog2b,log2blog3c,log3cloga are in A.P.
2(log2blog3c)=(logalog2b)+(log3cloga)
3log2b=3log3c2b=3c
Now, b2=acb2=a.2b3b=2a3,c=4a9
i.e., a=a,b=2a3,c=4a9
a:b:c=1:23:49=9:6:4
Since, sum of any two is greater than the 3rd,a,b,c form a triangle.

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