If a is the first term of a G.P.,l is the n^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $r$ be the common ratio of the $G.P.$The $n^{th}$ term is given by $l = ar^{n-1} \dots (1)$The first $n$ terms of the $G.P.$ are $a, ar, ar^2,

Vedclass · Accepted Answer

$(al)^{n/2}$

Vedclass · Answer

(A) Let $r$ be the common ratio of the $G.P.$The $n^{th}$ term is given by $l = ar^{n-1} \dots (1)$The first $n$ terms of the $G.P.$ are $a, ar, ar^2, \dots, ar^{n-1}$.The product $P$ is given by $P = a 	imes (ar) 	imes (ar^2) 	imes \dots 	imes (ar^{n-1})$.$P = a^n 	imes r^{1+2+3+\dots+(n-1)}$.Using the sum of the first $(n-1)$ natural numbers,$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$.$P = a^n 	imes r^{\frac{n(n-1)}{2}} = (a^2 	imes r^{n-1})^{n/2}$.Since $l = ar^{n-1}$,we can write $a^2 r^{n-1} = a(ar^{n-1}) = al$.Therefore,$P = (al)^{n/2}$.

If $a$ is the first term of a $G.P.$,$l$ is the $n^{th}$ term,and $P$ is the product of the first $n$ terms,then $P=$

Similar Questions

The sum of all natural numbers $n$ such that $100 < n < 200$ and $H.C.F. (91, n) > 1$ is

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms will be:

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + .....$ is

If $a$ is the arithmetic mean of $b$ and $c$,and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

If $s$ is the sum of an infinite $G.P.$ and $a$ is the first term,then the common ratio $r$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module