# Arithmetic Mean

Arithmetic mean of any two numbers â€˜*a*â€™ and â€˜

*b*â€™ is given by

When the arithmetic mean is placed between the numbers, the resulting numbers will be in A.P.

**Example 7.13:** The arithmetic mean of two numbers 10 and 14 is

We can see that, 10, 12 and 14 are in A.P. with a common difference â€˜2â€™.

We can conclude that,

If three numbers *a*, *b* and *c* are in A.P., then *b* is the arithmetic mean of *a* and *c* and *b*

Let *n* numbers *a*_{1}, *a*_{2} ... *a*_{n} be in A.P., then their arithmetic mean is given by

Example

Find the arithmetic mean of the progression 1, 3, 5, 7, 9.

Solution

Five numbers are given, so

Between two given numbers, it is possible to insert any number of terms such that the series thus formed shall be in A.P. The terms inserted are called the arithmetic means.

To insert

*n*= 5.Between two given numbers, it is possible to insert any number of terms such that the series thus formed shall be in A.P. The terms inserted are called the arithmetic means.

To insert

*n*arithmetic means between two given numbers*a*and*b*, we take the common difference as follows:Then,

*A*

_{1}=

*a*+

*d*,

*A*

_{2}=

*a*+ 2

*d*,

*A*

_{3}=

*a*+ 3

*d*

and so on up to

*A*=

_{n}*a*+

*nd*

Here,
The sum of

*A*_{1},*A*_{2},*A*_{3}, . . .*A*are the arithmetic means inserted between_{n}*a*and*b*.*n*arithmetic means between*a*and*b*is given by

Example

Insert three arithmetic means between 7 and 15.

Solution

Given

The common difference for inserting the arithmetic means are calculated as:
Using the above value of â€˜

To check whether the answers are correct or not, write down the arithmetic means that you calculated in between the two given numbers. If the resulting numbers are in A.P., then the arithmetic means you calculated are correct.
In the above example,

*a*= 7,*b*= 15 and*n*= 3.The common difference for inserting the arithmetic means are calculated as:

*d*â€™ , the three arithmetic means will be,*A*_{1}= 7 + 2 = 9*A*_{2}= 7 + 2(2) = 11*A*_{3}= 7 + 3(2) = 13*a*,*A*_{1},*A*_{2},*A*_{3},*b*= 7, 9, 11, 13, 15 respectively, which is an arithmetic progression with a common difference â€˜2â€™. Hence, we can conclude that our calculation is correct.