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


an = b = arn-1 —1


Here,


P = Product of n terms


P = (a) × (ar) × (ar2) × ….. × (arn-1)


P = (a × a × …a) × (r × r2 × …rn-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


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