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Question

If the sides of a triangle ABC are in A.P as well as in G.P, then the value of (r1r2r2r3) is (where r1,r2 and r3 are the radii of ex-circles):

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Solution

a,b,c are in A.P as well as in G.P.

b2=ac and a+c=2b

(a+c)2=4ac

(ac)2=0

a=c

2a=2b,a=b

r1=sa,r2=sb,r3=sc

Since a=b=c we have

r1=r2=r3

r1r2r2r3=0

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