Question

if a²,b²,c²are in A.P prove that a/b+c,b/c+a,c/a+b are in A.P
please solve this this is urgent​

Answer 1

Answer:

In AP form 2nd term - 1st term = 3rd term - 2nd term

b²-a² = c²-b²

b²+b² = c²+a²

2b² = c²+a²

Add 2ab+2ac+2bc on both sides

2b²+2ab+2ac+2bc = a²+c²+ac+ac+bc+bc+ab+ab

2b²+2ab+2ac+2bc = ac+bc+a²+ab+bc+c²+ab+ac

2b²+2ab+2ca+2cb = ca+cb+a²+ab+cb+c²+ab+ac

2(ba+b²+ca+cb) = (ca+cb+a²+ab) + (cb+c²+ab+ac)

2((ba+b²)+(ca+cb)) = ((ca+cb)+(a²+ab)) + ((cb+c²)+(ab+ac))

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

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

Divide whole by (a+b)(b+c)(c+a)

c/c+a = 1/b+c + 1/a+b

1/c+a + 1/c+a = 1/b+c + 1/a+b

1/c+a - 1/b+c = 1/a+b - 1/c+a

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2nd term - 1st term = 3rd term - 2nd term

Thus 1/b+c ,1/c+a ,1/a+b are in AP.

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