Q. 233.9( 76 Votes )

# 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 P^{2} = (ab)^{n}.

Answer :

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,

a_{1} = a = a

a_{n} = b = ar^{n-1} —1

Here,

P = Product of n terms

⇒ P = (a) × (ar) × (ar^{2}) × ….. × (ar^{n-1})

⇒ P = (a × a × …a) × (r × r^{2} × …r^{n-1})

⇒ P = a^{n} × 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 = a^{n} × r ^{1 + 2 +…+(n–1)}

⇒ P = a^{n} ×

⇒ P^{2} =

⇒ P^{2} =

⇒ P^{2} =

⇒ P^{2} =

⇒ P^{2} = from eq –1

∴ P^{2} = (ab)^{n}

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