raju
Last Activity: 9 Years ago
If a2, b2 and c2 are in A.P
b2 -a2 = c2-b2
(b-a)(b+a)=(c-b)(c+b)
(b-a)/(c+b)=(c-b)/(b+a)
(b-a+c-c)/(b+c)=(c-b+a-a)/(a+b)
(b-a+c-c)/(b+c)(c+a)=(c-b+a-a)/(a+b)(c+a)
{1/(c+a)} − {1/(b+c)} = {1/(a+b)} - {1/(c+a)}
1/(b+c),1/(c+a),1/(a+b) are in A.P.
So b+c, c+a, a+b are in H.P.