If a2,b2,c2 are in A.P to prove …
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Posted by Dev Saini 7 years, 1 month ago
- 2 answers
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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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.
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