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

Hint: 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.