Question

Hay guys I'm back after a long time.

Solve my this problem for 50 points.

If a, b, c are in A.P.
Then show that

a²(b+c), b²(c+a), c²(a+b) are in A.P.

Answer 1

a , b , c are in A. P
b - a = c - b constant
b^2(c+a) - a^2 (b+c) =(b-a)(ab+bc+ca)
C^2 (a+b)-b^2(c+a). = (c-b)(ab+bc+ca)

Difference is constant
a^2(b+c) , b^2(c+a) , c^2(a+b) are in A.P

Answer 2

_______✨ HEY MATE ✨______

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SOLUTION :-
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➡️a , b , c are in A. P
b - a = c - b constant
b^2(c+a) - a^2 (b+c) =(b-a)(ab+bc+ca)
C^2 (a+b)-b^2(c+a). = (c-b)(ab+bc+ca)

Difference is constant ⤵️
a^2(b+c) , b^2(c+a) , c^2(a+b) are in A.P.

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