Arithmetic Mean

 Arithmetic Mean

If three numbers are in arithmetic progression, then the middle term is called arithmetic mean of first and third term. Since the three numbers 2, 4, 6 are in arithmetic progression (with common difference d = 2), 4 is the arithmetic mean between 2 and 6.

If the three numbers a, b and c are in arithmetic progression, then by definition,

                b – a = c – b

 or,         b + b = a + c       

or,          2b = a + c

therefore, b = (a + c) / 2

Hence the arithmetic mean between a and c is (a + c) / 2.

Again, when any number of quantities are in A.P. all the terms in between the first term and the last term are called the arithmetic means between these two terms. For example, since, 5, 8, 11, 14, 17, 20, 23 are in A.P. so, 8, 11, 14, 17, 20 are the arithmetic means between 5 and 23. 

Arithmetic Means between two numbers a and b

Let a and b be two given numbers. Again, let m₁, m₂, m₃, …, m_n be n arithmetic means between a and b. Then a, m₁, m₂, m₃, …, m_n, b is in A.P.

Here, number of arithmetic means = n

So, number of terms of A.P. = n + 2

It means,

                b             = (n + 2) ^th term of A.P.

or,          b             = a + (n + 2 – 1) d, where d is common difference

or,          b             = a + (n + 1) d

or,          (n + 1) d               = b – a

therefore, d       = (b – a) / (n + 1)

Now,     arithmetic means are given by,

                m₁          = a + d                   = a + (b – a) / (n + 1)

                m₂          = a + 2d                = a + 2 (b – a) / (n + 1)

                m₃          = a + 3d                = a + 3 (b – a) / (n + 1)

                .

                .

                .

                m_n          = a + nd                = a + n (b – a) / (n + 1)

1.  Find the A.M. between 2 and 4.

Solution,

Let ‘m’ be the A.M. between 2 and 4,

So,          a = 2, b = 4, m =?

We have,

A.M.      = (a + b) / 2

or, m     = (2 + 4) / 2         = 3

2.  If 3, x, y, -9 are in A.P., find the value of x and y.

Solution,

Here,     first term (a) = 3

                Number of terms (n) = 4

                Last term (t₄) = -9

i.e.          a + (n – 1) d = -9

or,          3 + (4 – 1) d = -9

                3d = − 9 – 3

or,          d             = - 12 / 3              = − 4

Now,     x = a + d = 3 + (-4) = 3 – 4 = -1

                y = a + 2d = 3 + 2 (-4) = 3 – 8 = -5  

3.  Insert 5 arithmetic means between 11 and 29.

Solution,

Here,     first term (a) = 11,

                last term (b) = 29,

                number of means (n) = 5

Therefore,          d = (b – a) / n + 1              = (29 – 11) / (5 + 1)          =         = 3

Now,     m₁          = a + d   = 11 + 3                = 14

                m₂          = a + 2d = 11 + 6                = 17

                m₃          = a + 3d = 11 + 9                                = 20

                m₄          = a + 4d = 11 + 12                              = 23

                m₅          = a + 5d = 11 + 15                              = 26

Second method,

Here,     first term (a) = 11,

number of terms (n) = 5 + 2 = 7

last term (t₇)      = 29

i.e.          a + 6d    = 29

or,          11 + 6d = 29

or,          6d           = 29 – 11

or,          d             =         = 3

Now,     m₁          = a + d   = 11 + 3                 = 14

                m₂          = a + 2d = 11 + 6                 = 17

                m₃          = a + 3d = 11 + 9                 = 20

                m₄          = a + 4d = 11 + 12              = 23

                m₅          = a + 5d = 11 + 15              = 26

4.  The product of two numbers is 45 and their arithmetic means is 7. Find the numbers.

Solution,

Let a and b be the two required numbers. Then,

                a. b         = 45

or,          b             =  ------------------(i)

Again,   (a + b) / 2             = 7

or,          a + b      = 14 -----------------(ii)

putting the value of b in (ii), we get,

                a + 45 / a = 14

or,          a² + 45 = 14 × a

or,          a² – 14a + 45 = 0

or,          a² – 9a – 5a + 45 = 0

or,          (a – 9) (a – 5)      = 0

therefore,           a              = 5 or 9 

If a = 5, then b   = 45 / a = 45 / 5 = 9

If a = 9, then b   = 45 / 9                 = 5

Hence, the required two numbers are 5 and 9 or 9 and 5.

5.  There are n arithmetic means between 20 and 80 such that the ratio of the first mean to the last mean is 1:3. Find n.

Solution,

Here, the sequence is 20, ……., 80.

Since there are n A.M.’s the total number of terms = n + 2 (including the first and the last term)

Since, they are in A.P. we can write 20, 20 + d, ……., 80 – d, 80

Where, 20 + d is the second term or first mean and 80 – d is second last term of last mean.

By the question,

First mean / last mean = 1 / 3

or,          (20 + d) / (80 – d) = 1 / 3

or,          60 + 3d = 80 – d

or,          3d + d    = 80 – 60

or,          4d           = 20

or,          d             = 20 / 4 = 5

Now we know,

                t_n             = a + (n – 1) d

or,          80           = 20 + (n – 1) 5

or,          60           = 5n – 5

or,          65           = 5n

or,          n             = 65 / 5

or,          n             = 13

There are n – 2 = 13 – 2 = 11 A.M.’s between 20 and 80.