Question

if a, b, c are in AP then prove that a square bracket of B + c, b square bracket of C plus a, c square bracket of a + b are also in A. p.

Answer 1

A , b , c are in AP
so, common difference must be constant.
b - a = c - b
now, multiply with (ab + bc + ca) in both sides,
(b - a)(ab + bc + ca) = (c - b)(ab + bc + ca)
(b - a)[ab + c(b + a)] = (c - b)[bc + a(c + b)]
(b - a)ab + c(b + a)(b - a) = (c - b)bc + a(c + b)(c - b)
(ab² - a²b) + c(b² - a²) = (c²b - bc²) + a(c² - b²)
(ab² - a²b) + (cb² - a²c) = (c²b - b²c) + (ac² - ab²)
(ab² + cb²) -(a²b + a²c) = (c²b + ac²) - (bc² + ac²)


you can observed that ,
it seems like a²(b + c) , b²(c + a), c²(a + b) are in AP