If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.

Asked by Abhisek | 1 year ago |  79

##### Solution :-

Given, a, b, c,and d are in G.P.

So, we have

b2 = ac … (i)

c2 = bd … (ii)

Required to prove (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,

(bn + cn)2 = (an + bn) (cn + dn)

Taking L.H.S.

(bn + cn)2 = b2n + 2bncn + c2n

= (b2)n+ 2bncn + (c2) n

= (ac)n + 2bncn + (bd)n  [Using (i) and (ii)]

= an cn + bncn+ bn cn + bn dn

= an cn + bncn+ an dn + bn dn  [Using (iii)]

= cn (an + bn) + dn (an + bn)

= (an + bn) (cn + dn)

= R.H.S.

Therefore, (an + bn), (bn + cn), and (cn + dn) are in G.P

Hence proved.

Answered by Pragya Singh | 1 year ago

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