Question

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 .

Answer 1

The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar2 , ar3 , … arn-1, where r is the common ratio.
b = arn-1 … (1)
P = Product of n terms
= (a) (ar) (ar2 ) … (arn-1)
= (a × a ×…a) (r × r 2 × …rn-1)
= an r1 + 2 +…(n-1) … (2)
Here, 1, 2, …(n-1) is an A.P.

$∴1 + 2 + ……….+ (n-1) =\frac{ n - 1 }{ 2} [2 + (n - 1 - 1) × 1] =\frac{ n - 1 }{ 2} [2 + n - 2] = \frac{n(n - 1)} { 2}$

$P = a^n r\frac{n(n - 1) }{ 2}$
$∴ P^2 = a^{2n }\,r^{n(n - 1)}$
= [a^2 r^{(n - 1)}]^n$$ = [ a × a^{rn - 1}]^n
$= (ab)^n$ $[Using\,\,\, (1)]$
Thus, the given result is proved.