Question

a²,b²,c² in A.P prove that 1/b+c,1/c+a,1/a+b are A.P .​

Answer 1

Answer:

Here is your answer hope help you

Step-by-step explanation:

2b2 = a2 + c2 [as a2 b2 c2 are in A.P.]

=> b2 + b2 = a2 + c2

=> b2 - a2 = c2 - b2

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

=> (b−a)(c+b)=(c−b)(b+a)

Divide both side by 1(c+a), We get

=> (b−a)(c+b)×(c+a)=(c−b)(b+a)×(c+a)

=> (b−a+c−c)(c+b)×(c+a)=(c−b+a−a)(b+a)×(c+a)

=> (b+c)−(c+a)(c+b)×(c+a)=(c+a)−(a+b))(b+a)×(c+a)

=> 1(c+a)−1(c+b)=1(a+b)−1(c+a)

=> 2(c+a)=1(a+b)+1(c+b)

Hence by this equation we, can say that 1b+c,1c+a,1a+bare in A.P.