# Insertion of Arithmetic Mean (AM) between Two Numbers

## Insertion of Arithmetic Mean (AM) between Two Numbers

Arithmetic progression (AP): A sequence is said to be an arithmetic progression, if the difference of a term and previous term is always same.

i, e.., an₊₁ – an = constant (d)

The constant difference, generally denoted by ‘d’ is called the common difference.

Example:

(i) 1, 2, 5, 7, …

(ii) 2, 4, 6, …

The nth term of an AP: Let ‘a’ be the first term ‘d’ be the common difference and l be the last term of an AP, then nth term is given by

Tn = l = a + (n – 1) d

The nth term from last is T’n = l – (n – 1) d.

The sum of n terms of an AP: Suppose there are n terms of a sequence, whose first term is ‘a’, common difference is d and last term is I, then sum of the n terms is given by Sn = n/2 [2a + (n – 1) d].

Insertion of Arithmetic Mean (AM) between Two Numbers: Let A₁, A₂, …, An, n arithmetic means are inserted between two numbers ‘a’ and ‘b’ such that a, A₁, A₂, …, An, b from an AP.

Here, total number of terms are (n + 2) and common difference be d

b = (n + 2)th term = a + (n + 2 – 1) d

d = (b – a)/ (n + 1)

$${{A}_{1}}=a+d=a+\left( \frac{b-a}{n+1} \right)=\frac{na+b}{n+1}$$,

$${{A}_{2}}=a+2d=a+2\left( \frac{b-a}{n+1} \right)=\frac{(n-1)a+2b}{n+1}$$,

$${{A}_{r}}=a+r\left( \frac{b-a}{n+1} \right)=\frac{(n-r+1)a+rb}{n+1}$$,

If r = 1 and n = 1, then

$${{A}_{1}}=\frac{(1-1+1)a+(1)b}{1+1}$$,

$${{A}_{1}}=\frac{a+b}{2}$$ = Mean.