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Question
If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.
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Solution
It is given that a, b, c,and d are in G.P.
∴b2 = ac … (1)
c2 = bd … (2)
ad = bc … (3)
It has to be proved that (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,
(bn + cn)2 = (an + bn) (cn + dn)
Consider L.H.S.
(bn + cn)2 = b2n + 2bncn + c2n
= (b2)n+ 2bncn + (c2) n
= (ac)n + 2bncn + (bd)n [Using (1) and (2)]
= an cn + bncn+ bn cn + bn dn
= an cn + bncn+ an dn + bn dn [Using (3)]
= cn (an + bn) + dn (an + bn)
= (an + bn) (cn + dn)
= R.H.S.
∴ (bn + cn)2 = (an + bn) (cn + dn)
Thus, (an + bn), (bn + cn), and (cn + dn) are in G.P.
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