If a2,b2,c2 are in A.P to prove …

If a², b², c² are in AP, then

2b² = a² + c² 

b² + b² = a² + c² 

b² - a² = c² - b² 

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

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

Dividing both sides by (c + a),

(b - a)/{(c + b)(c + a)} = (c - b)/{(b + a)(c + a)}

{(b + c) - (c + a)}/{(b + c)(c + a) = {(c + a) - (a + b)}/{(a + b)(c + a)}

{1/(c + a)} - {1/(b + c)} = {1/(a + b)} - {1/(c + a)}

{2/(c + a)} = {1/(a + b)} - {1/(b + c)}

Therefore 1/(a + b), 1/(b + c), 1/(c + a) are in AP.

Ans. Given : a² b² c² are in A.P.

To Prove : ${1 \over b +c } ,{1 \over c+a}, {1\over a+b}$ Are in A.P.

Proof : 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)

=> ${(b-a) \over (c+b) } = {(c-b) \over (b+a)}$

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

=> ${(b-a) \over (c+b) \times (c+a) } = {(c-b) \over (b+a)\times(c+a)}$

=> ${(b-a+c-c) \over (c+b) \times (c+a) } = {(c-b+a-a) \over (b+a)\times(c+a)}$

=> ${(b+c) - (c+a) \over (c+b) \times (c+a) } = {(c+a) -(a+b)) \over (b+a)\times(c+a)}$

=> ${1 \over(c+a)} - {1\over (c+b) } = {1\over (a+b)} -{ 1\over(c+a)}$

=> ${2 \over(c+a)} = {1\over (a+b)} + {1\over (c+b) }$

Hence by this equation we, can say that ${1 \over b +c } ,{1 \over c+a}, {1\over a+b}$are in A.P.