a,b&c are in A.P . Prove that b+c, c+a & a+b are in A.P.
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If a , b, c are in AP,
b-a = c-b [common difference are same]
b-a +c-c=c-b+a-a
(b+c)-(a+c)=(c+a)-(a+b)
(a+b)+(b+c)= 2(c+a)
i.e.a+b, a+c. b+a are in AP
(b+c) -(a+c) = (c+a) -(a+b)
by changing the arrangement
ok
(b+c) + (a+b) = 2(a+c)
so. b+c, a+b and a+c are in AP
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