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If a2,b2,c2 are in A.P to prove …

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

  • 2 answers

Rashmi Bajpayee 7 years, 1 month ago

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.

Naveen Sharma 7 years, 1 month ago

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.

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