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Question

Question: How do you find the geometric mean between 81 and 4?...

How do you find the geometric mean between 81 and 4?

Explanation

Solution

First try to figure out the number of terms given to you whose geometric mean is to be found out. After this step you just have to use the formula for finding out geometric mean which consists of multiplication of all the terms and taking the nth{{\text{n}}^{{\text{th}}}} root where n{\text{n}} denotes the number of terms.

Formula used:
Let n{\text{n}} be the number of terms whose geometric mean is to be found and let x1,x2,x3...xn{x_1},{x_2},{x_3}...{x_{\text{n}}} be the n{\text{n}} terms given to us whose geometric mean is to be found then geometric mean is given by the formula x1.x2.x3...xnn\sqrt[{\text{n}}]{{{x_1}.{x_2}.{x_3}...{x_{\text{n}}}}} respectively.

Complete step by step solution:
Starting with the definition of geometric mean, Geometric mean is nothing but nth{{\text{n}}^{{\text{th}}}} root of the multiplication of n{\text{n}} terms.
That is suppose if we are told to find the geometric mean of n{\text{n}} terms we can simply solve it by the formula of geometric mean which is x1.x2.x3...xnn\sqrt[{\text{n}}]{{{x_1}.{x_2}.{x_3}...{x_{\text{n}}}}} where x1,x2,x3...xn{x_1},{x_2},{x_3}...{x_{\text{n}}} denote n{\text{n}} terms and hence we take nth{{\text{n}}^{{\text{th}}}} root.
Now in the problem given to us, we have been told to find out the geometric mean between two numbers 81 and 4 respectively.
Compared to above general formula, our number of terms is 2 i.e. n = 2{\text{n = 2}} and x1=81{x_1} = 81 , x2=4{x_2} = 4 .
Hence substituting in the above general formula we get
Geometric mean = x1.x22\sqrt[2]{{{x_1}.{x_2}}}
=81×42 =3242 =18  = \sqrt[2]{{81 \times 4}} \\\ = \sqrt[2]{{324}} \\\ = 18 \\\
Hence, we have obtained the geometric mean of our two numbers 81 and 4 as 18.

Note: Always remember that while calculating the geometric mean the thing we tend to forget is the nth{{\text{n}}^{{\text{th}}}} root where n{\text{n}} denotes the number of terms whose geometric mean is to be found out. Often we neglect the nth{{\text{n}}^{{\text{th}}}} root part and consider it as square root which can make us go completely wrong with our solution.