An **arithmetic progression** is a sequence of numbers such that the difference of any two successive members is a constant.

For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference **1**.

Second example: the sequence 3, 5, 7, 9, 11,... is an arithmetic progression

with common difference **2**.

Third example: the sequence 20, 10, 0, -10, -20, -30, ... is an arithmetic progression

with common difference **-10**.

#### Notation

We denote by

**d**the common difference.

By

**a**we denote the

_{n}**n**-th term of an arithmetic progression.

By

**S**we denote the sum of the first n elements of an arithmetic series.

_{n}**Arithmetic series**means the sum of the elements of an arithmetic progression.

#### Properties

_{1}+ a

_{n}= a

_{2}+ a

_{n-1}= ... = a

_{k}+ a

_{n-k+1}

and

_{n}= ½(a

_{n-1}+ a

_{n+1})

Sample: let 1, 11, 21, 31, 41, 51... be an arithmetic progression.

51 + 1 = 41 + 11 = 31 + 21

and

11 = (21 + 1)/2

21 = (31 + 11)/2...

If the initial term of an arithmetic progression is a

_{1}and the common difference of successive members is d, then the n-

*th*term of the sequence is given by

_{n}= a

_{1}+ (n - 1)d, n = 1, 2, ...

The sum S of the first n numbers of an arithmetic progression is given by the formula:

_{1}+ a

_{n})n

where a

_{1}is the first term and a

_{n}the last one.

or

_{1}+ d(n-1))n

#### Arithmetic Progression Problems

1) Is the row 1,11,21,31... an arithmetic progression?

**Solution:**Yes, it is an arithmetic progression. Its first term is 1 and the common differnece is 10.

2) Find the sum of the first 10 numbers of this arithmetic series: 1, 11, 21, 31...

**Solution:**we can use this formula S =

^{1}/

_{2}(2a

_{1}+ d(n-1))n

S =

^{1}/

_{2}(2.1 + 10(10-1))10 = 5(2 + 90) = 5.92 = 460

3) Try to prove that if the numbers 1/(c + b), 1/(c + a), 1/(a + b) form an arithmetic progression then the numbers a

^{2}, b

^{2}, c

^{2}form an arithmetic progression too.