Answer :

Given: a, b and c are in A.P.

Now, for a, b, c to be in A.P, their common difference should be same. Therefore,

b - a = c - b

∴ a + c = 2b --------- (1)

**To prove:** b + c, c + a and a + b are in A.P.

Common difference will remain same,

(c + a) - (b + c) = (a + b) - (c + a)

then, (a + b) + (b + c) = 2(a + c)

i.e. now we have to prove, (a + b) + (b + c) = 2(a + c) ----------- (2)

L.H.S of equation (2)

(a + b) + (b + c)

= a + 2b + c

from equation (1), put 2b = a + c

= a + a + c + c

= 2a + 2c

= 2(a + c) --- RHS

**Hence, it is proved that (b + c), (c + a) and (a + b) are in AP.**

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