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Discussion Response Post to:
A.P, G.P, H.P
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Algebra
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Author
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22 Apr 2007 20:34:24 IST
Subject:
Re:A.P, G.P, H.P
Avinash_Bhat
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3)
1/(b - a) + 1/(b - c) = 1/a + 1/c
ie, 1/(b - a) - 1/c = 1/a - 1/(b - c)
=> (c - b + a)/(bc - ac) = (b - c - a)/(ab - ac)
=> (a - b + c)/(bc - ac) = - (a - b + c)/(ab - ac)
=> bc - ac = ac - ab
=> b = 2ac/(a + c)
=> a, b, c in H.P.
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