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

Solution:

It is given that a, b, c, d are in G.P.

Therefore,

b2 = ac ....(1)

c2 = bd ....(2)

ad = bc ....(3)

We need to prove (an + bn), (bn + cn), (cn + d n) are in G.P. i.e., we have to prove that

(bn + cn)2 = (an + bn)×(cn + dn)

Consider,

(bn + cn)2 = b2n + 2bncn + c2n

= (b2)n + 2bncn + (c2)n

= (ac)n + 2bncn + (bd)n [Using (1) and (2)]

= ancn + bncn + bncn + bndn

= ancn + bncn + andn + bndn [Using (3)]

= cn (an + bn) + dn (an + bn)

= (an + bn)(cn + d n)

Hence, (bn + cn)2 = (an + bn)×(cn + dn)

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

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 17

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

Summary:

If, a, b, c, d are in G.P. we proved that (an + bn), (bn + cn), (cn + d n) are in G.P

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