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

Given: Let the first and the nth term of a G.P. be a and b, respectively, and P be the product of n terms.

Here,

a₁ = a = a

a_n = b = ar^n-1 —1

Here,

P = Product of n terms

⇒ P = (a) × (ar) × (ar²) × ….. × (ar^n-1)

⇒ P = (a × a × …a) × (r × r² × …r^n-1)

⇒ P = aⁿ × r ^{1 + 2 +…(n–1)} —2

Here,

1, 2, …(n – 1) is an A.P.

The sum of n terms of an A.P is given by : (here n: no of terms, a:first term, d:common difference)

∴ 1+2+3+….+(n-1) =

∴ 1+2+3+….+(n-1) =

∴ P = aⁿ × r ^{1 + 2 +…+(n–1)}

⇒ P = aⁿ ×

⇒ P² =

⇒ P² =

⇒ P² =

⇒ P² =

⇒ P² = from eq –1

∴ P² = (ab)ⁿ