If a2 , b2 , c2 are in AP then prove that 1 b+c , 1 c+a , 1 a+b are in AP.
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Step-by-step explanation:
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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