If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2

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

Thus,

The G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.

Then,

b = ar^n–1  … (1)

P = Product of n terms

= (a) (ar) (ar²) … (ar^n–1)

= (a × a ×…a) (r × r² × …r^n–1)

= aⁿ 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,

p²= (ab)ⁿ

Therefore, p²= (ab)ⁿ is proved.