Given, a, b, c,and d are in G.P.
So, we have
b2 = ac … (i)
c2 = bd … (ii)
ad = bc … (iii)
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 agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).