if a,b,c are in H.P , then
2/b = 1/a +1/c --------------{1}
= (a+c)/ac
2ac = ab+bc
ac = (ab+bc)/2 --------------{2}
And Given a2 , b2 , c2 are in H.P.
Therefore , 2/b2 = 1/a2 +1/c2
= (a2+c2)/a2c2
2a2c2 = a2b2+b2c2
a2c2 = (a2b2 + b2c2)/2 -------------{3}
{(ab+bc)/2}2 = (a2b2+b2c2)/2
(a2b2 + b2c2 + 2ab2c)/4 = (a2b2 + b2c2)/2
2ab2c = a2b2 + b2c2
(a2b2 + b2c2) – 2(ab)(bc) = 0
(ab-bc)2 = 0
ab= bc
a =c ------------{4}
From {1} ,
2/b = 1/a+ 1/c
= 1/a +1/a
= 2/a
a=b ----------------{5}
From {4} , {5} , we get,
a=b=c.