The 5^{th} term of a G.P. is 2, find the prod | vedclass

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

Question

(B) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.Given that the $5^{

Vedclass · Accepted Answer

$512$

Vedclass · Answer

(B) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.Given that the $5^{th}$ term is $2$, we have:$a_5 = a r^{5-1} = a r^4 = 2$ $...(1)$We need to find the product of the first $9$ terms:$P = a 	imes (a r) 	imes (a r^2) 	imes \dots 	imes (a r^8)$$P = a^9 	imes r^{(1 + 2 + 3 + \dots + 8)}$The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.For $n=8$, the sum is $\frac{8 	imes 9}{2} = 36$.Therefore, $P = a^9 	imes r^{36} = (a r^4)^9$.Substituting the value from equation $(1)$:$P = (2)^9 = 512$.

The $5^{th}$ term of a $G.P.$ is $2,$ find the product of first $9$ terms.

Similar Questions

$A$ person has two parents (father and mother),four grandparents,eight great-grandparents,and so on. Find the total number of ancestors the person has up to the $10^{th}$ generation.

If the product of three consecutive terms of a $G.P.$ is $216$ and the sum of their products taken two at a time is $156$,then the numbers are:

The sum of all the digits of the numbers from $1$ to $100$ is

Given $a_1, a_2, a_3, \dots$ form an increasing geometric progression with common ratio $r$ such that $\log_8 a_1 + \log_8 a_2 + \dots + \log_8 a_{12} = 2014$,then the number of ordered pairs of integers $(a_1, r)$ is equal to

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}, (x \ne 0)$,then $a, b, c, d$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module