Answers

2015-05-05T14:36:32+05:30
Let a=3
3,b,c are in AP, so b-3=c-b\implies c=2b-3

a+b+c=25
substituting a=3 and c=2b-3 
3+b+2b-3=25
b=\frac{25}{3}
\implies c=\frac{41}{3}

Thus the required numbers are 3,~\frac{25}{3},~\frac{41}{3}
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