Question

Geometric mean of $7,{7^2},{7^3},...,{7^n}$ is
A. ${7^{\dfrac{{(n + 1)}}{2}}}$
B. $7$
C. ${7^{\dfrac{n}{2}}}$
D. ${7^n}$

Answer 1

Here we are asked to find the geometric mean of the given geometric progression. An average value that represents the central tendency of the given data is called a geometric mean. To find the geometric mean of a geometric progression we need to find the ${n^{th}}$ root of that product of all the terms in that geometric progression.

Complete step-by-step solution:
Given geometric progression $7,{7^2},{7^3},...,{7^n}$ , we aim to find the geometric mean of this geometric progression.
We know that to find the geometric mean of a geometric progression we need to find the ${n^{th}}$ root of that product of all the terms in that geometric progression.
First, let us find the product of all the terms in the given geometric progression.
If the series ${a_1},{a_2},{a_3},...,{a_n}$ is a geometric progression, then the product of all of the terms will be ${a_1}.{a_2}.{a_3}...{a_n}$.
Here the geometric progression is ${a^x}.{a^y} = {a^{x + y}}$ then the product of all of its terms will be ${7.7^2}{.7^3}{....7^n}$.
Form exponents and powers, if the base value is the same in the product, then we can add their powers.
That is ${a^x}.{a^y} = {a^{x + y}}$ . Thus, we get
${7.7^2}{.7^3}{....7^n} = {7^{\left( {1 + 2 + 3 + ... + n} \right)}}$
$= {7^{\dfrac{{n(n + 1)}}{2}}}$
Thus, we have found the product of all the terms in the given geometric progression.
Now let us find the geometric mean by taking ${n^{th}}$ root to the above product value.
Geometric progression of 7,{7^2},{7^3},...,{7^n}$$$$ = \sqrt[n]{{{7^{\dfrac{{n(n + 1)}}{2}}}}}
We know that on taking square root for a squared number, the power two and the square root will get canceled. That is $\sqrt {{a^2}} = a$ likewise on taking ${n^{th}}$ root for a number raised to the power$n$, the root and the power $n$ will get canceled. Thus, we get
Geometric progression of $7,{7^2},{7^3},...,{7^n} = \sqrt[n]{{{7^{\dfrac{{n(n + 1)}}{2}}}}} = {7^{\dfrac{{(n + 1)}}{2}}}$
Thus, we have found the geometric mean of the given geometric progression $7,{7^2},{7^3},...,{7^n}$is${7^{\dfrac{{(n + 1)}}{2}}}$.
Now let us see the options, option (a) ${7^{\dfrac{{(n + 1)}}{2}}}$is the correct option since we got the same value in our calculation above.
Option (b) $7$is an incorrect answer as we got ${7^{\dfrac{{(n + 1)}}{2}}}$as the correct answer in our calculation.
Option (c) ${7^{\dfrac{n}{2}}}$is an incorrect answer as we got ${7^{\dfrac{{(n + 1)}}{2}}}$as the correct answer in our calculation.
Option (d) ${7^n}$is an incorrect answer as we got ${7^{\dfrac{{(n + 1)}}{2}}}$as the correct answer in our calculation.

Note: Here we have used the generalization of the sum of $n$natural numbers. The sum of first $n$ natural numbers $1,2,3,...,n$can be generalized to$\dfrac{{n(n + 1)}}{2}$. That is $1 + 2 + 3 + ... + n = \dfrac{{n(n + 1)}}{2}$. The geometric mean of a geometric progression can also be calculated by taking the square root of the product of the first and the last term of the series that is$\sqrt {{a_1}.{a_n}}$. Unlike other means, geometric mean is more precise when there is more instability in the data.