Let’s take A to be the first term and R to be the common ratio of the G.P.
Then according to the question, we have
ARp–1 = a
ARq–1 = b
ARr–1 = c
Then,
aq–r br–p cp–q
= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r–p) × Ap–q × R(r –1)(p–q)
= Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)
= A0 × R0
= 1
Hence proved.
Answered by Abhisek | 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 \( \dfrac{81}{2}\).