# Arithmetic Mean

- The A.M. between two numbers a and b is

A =(a +b)/2 - Similarly, if a, A
_{1}, A_{2}, ..., A_{n}, b are in A.P., then A_{1}, A_{2}, ..., A_{n}are called n**arithmetic means**between a and b.

## Illustrative Examples

### Example

Insert six arithmetic means between 2 and 16. Also prove that their sum is 6 times the A.M. between 2 and 16.

### Solution

Let A_{1}, A_{2}, ..., A_{6} be six A.M.s between 2
and 16. Then, by def., 2, A_{1}, A_{2}, ..., A_{6}, 16
are in A.P.

Let d be the common difference. Here 16 is the 8th term.

16 = 2 +7d => d = 2.

Hence six A.M.s are a +d, a +2d, ..., a +6d

i.e. 4, 6, 8, 10, 12, 14.

Now sum of these means

= 6(4 +14)/2 = 54 = 6.9 = 6(2 +16)/2

= 6 times the A.M. between 2 and 16.

### Example

If A is the A.M. between a and b, show that

(A +2 a)/(A -b) + (A +2 b)/(A -a) = 4

### Solution

Since A is the A.M. between a and b, A = (a +b)/2

Hence (A +2 a)/(A -b) + (A +2 b)/(A -a)

=

= (5a + b)/(a - b) + (5b + a)/(b - a) = (4a - 4b)/(a-b)

= 4

## Exercise

- (i) Insert 5 arithmetic means between -2 and 10. Show that their sum is 5
times the arithmetic mean between -2 and 10.

(ii) Insert 4 arithmetic means between 12 and -3.

(iii) Insert 10 A.M.s between -5 and 17 and prove that their sum is 10 times the A.M. between -5 and 17. - If A is the A.M. between a and b, prove that

(i) (A -a)² +(A -b)² = (a -b)²/2

(ii) 4(a -A)(A -b) = (a -b)². - The ratio of the first to the last of n A.M.s between 5 and 35 is 1 : 4. Find n.
- There are n arithmetic means between 3 and 17. The ratio of the first mean to the last mean is 1 : 3. Find n.
- If the A.M. between p th and q th items of an A.P. is equal to the A.M. between rth and sth items, show that p +q = r +s.

## Answers

**1.**(i) 0, 2, 4, 6, 8 (ii) 9, 6, 3, 0

(iii) -3, -1, 1, 3, 5, 7, 9, 11, 13, 15

**3.**n = 9

**4.**n = 6