Given, the first term of the G.P is a and the last term is b.

Thus,

The G.P. is a, ar, ar^{2}, ar^{3}, … ar^{n–1}, where r is the common ratio.

Then,

b = ar^{n–1} … (1)

P = Product of n terms

= (a) (ar) (ar^{2}) … (ar^{n–1})

= (a × a ×…a) (r × r^{2} × …r^{n–1})

= a^{n }r ^{1 + 2 +…(n–1)} … (2)

Here, 1, 2, …(n – 1) is an A.P.

= \( \dfrac{n-1}{2}[2+(n-1-1)\times 1]\)

= \( \dfrac{n-1}{2}[2+n-2]\)

= \( \dfrac{n(n-1)}{2}\)

And, the product of n terms P is given by,

\( p=a^nr^\dfrac{n(n-1)}{2}\)

\( p^2=a^{2n}r^{n(n-1)}\)

\( p^2=[a^2r^{(n-1)}]^n\)

\([a\times ar^{n-1}]^n\)

Substituting (1) in the equation,

p^{2}= (ab)^{n}

Therefore, p^{2}= (ab)^{n }is proved.

Construct 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 \( \dfrac{81}{2}\).