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

AR^{p–1 }= a

AR^{q–1 }= b

AR^{r–1 }= c

Then,

a^{q–r }b^{r–p }c^{p–q}

= A^{q–r }× R^{(p–1) (q–r)} × A^{r–p} × R^{(q–1) (r–p)} × A^{p–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)}

= A^{0} × R^{0}

= 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) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).