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Question

Mathematics Question on nth Term of an AP

If loga,logb\log a, \log b, and logc\log c are in A.P. and also logalog2b,log2blog3c,log3cloga\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

a, b, c are the sides of a triangle

Explanation

Solution

loga,logb,logc\log a, \log b, \log c are in A.P.
2logb=loga+logc\Rightarrow 2 \log b =\log a +\log c
logb2=log(ac)\Rightarrow \log b ^{2}=\log ( ac )
b2=aca,b,c\Rightarrow b ^{2}= ac \Rightarrow a , b , c are in G.P.
logalog2b,log2blog3c,log3cloga\log a -\log 2 b , \log 2 b -\log 3 c , \log 3 c -\log a are in A.P.
2(log2blog3c)=(logalog2b)+(log3cloga)\Rightarrow 2(\log 2 b -\log 3 c )=(\log a -\log 2 b )+(\log 3 c -\log a )
3log2b=3log3c2b=3c\Rightarrow 3 \log 2 b =3 \log 3 c \Rightarrow 2 b =3 c
Now, b2=acb^{2}=a c
b2=a2b3\Rightarrow b^{2}=a \cdot \frac{2 b}{3}
b=2a3,c=4a9\Rightarrow b=\frac{2 a}{3}, c=\frac{4 a}{9}
i.e., a=a,b=2a3,c=4a9 a=a, b=\frac{2 a}{3}, c=\frac{4 a}{9}
a:b:c=1:23:49=9:6:4\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,c3^{ rd }, a , b , c form a triangle.