The geometric mean of the sequence 1, 2, 2^2, | vedclass

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

Question

(A) The sequence is $1, 2, 2^2, ..., 2^n$. The total number of terms is $n+1$.The geometric mean $GM$ is given by the $(n+1)$-th root of the product o

Vedclass · Accepted Answer

$2^{\frac{n}{2}}$

Vedclass · Answer

(A) The sequence is $1, 2, 2^2, ..., 2^n$. The total number of terms is $n+1$.The geometric mean $GM$ is given by the $(n+1)$-th root of the product of the terms:$GM = (1 	imes 2 	imes 2^2 	imes ... 	imes 2^n)^{\frac{1}{n+1}}$Using the property of exponents,the product is:$1 	imes 2^1 	imes 2^2 	imes ... 	imes 2^n = 2^{0+1+2+...+n}$The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$,so:$2^{0+1+2+...+n} = 2^{\frac{n(n+1)}{2}}$Now,substitute this back into the $GM$ formula:$GM = (2^{\frac{n(n+1)}{2}})^{\frac{1}{n+1}} = 2^{\frac{n(n+1)}{2(n+1)}} = 2^{\frac{n}{2}}$

The geometric mean of the sequence $1, 2, 2^2, ..., 2^n$ is:

Similar Questions

What will $Rs. 500$ amount to in $10$ years after its deposit in a bank which pays an annual interest rate of $10\%$ compounded annually?

If $\log_x a, a^{x/2}$ and $\log_b x$ are in $G.P.$,then $x = $

If $1 + \cos \alpha + \cos^2 \alpha + \dots \infty = 2 - \sqrt{2}$,then $\alpha$ $(0 < \alpha < \pi)$ is

If $r > 1$,$x = a + \frac{a}{r} + \frac{a}{r^2} + \dots \infty$,$y = b - \frac{b}{r} + \frac{b}{r^2} - \dots \infty$,and $z = c + \frac{c}{r^2} + \frac{c}{r^4} + \dots \infty$,then $\frac{xy}{z} = \dots$

The arithmetic mean of two numbers $b$ and $c$ is $a$,and $g_1$ and $g_2$ are two geometric means between them. If $g_1^3 + g_2^3 = kabc$,then $k = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module