Progression is a sequence of numbers in a particular order. The arithmetic progression is the most commonly used sequence with easy to understand formulas.

If we observe in our regular lives, we come across progression quite often. Roll numbers of our class, days in a week or months in a year. Did you notice that counting numbers, even numbers or odd numbers, all follow a particular progression!

A progression is a special type of sequence for which it is possible to obtain a formula for the \( n^{th} \)

Example:

11, 23, 35, 47, …

9,9,9,9,…

26, 13, 0, -13,…

5, 25, 125, 625,…

## Arithmetic Progression:

In all of the examples given for progressions except the last one, we observe that every term (excluding the first term) is obtained by adding a fixed number to the previous terms. That makes all those progressions arithmetic. We will refer arithmetic progression with the abbreviation AP from now on.

Definition 1: An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.

Definition 2: The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.

Consider an AP to be

\(a_{1}, a_{2}, a_{3}, ……..a_{n}\)

So, \(a_{2} – a_{1} = a_{3} – a_{2} = ……. = a_{n} – a_{n-1} = d\)

where d is the common difference.

The progression can also be written in terms of common difference, as follows

a, a+d, a+2d, a+3d, a+4d, ………. a+(n-1)d

where a is the first term of the progression and d is the common difference.

Position of Terms |
Representation of Terms |
Values of Term |

1 |
\(a_{1}\) |
a + d = a+(1-1) + d |

2 |
\(a_{2}\) |
a + 2d = a+(2-1) + d |

3 |
\(a_{3}\) |
a + 3d = a+(3-1) + d |

4 |
\(a_{4}\) |
a + 4d = a+(4-1) + d |

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n |
\(a_{n}\) |
a + (n-1)d |

## nth Term of an AP

For a given AP, where a is the first term, d is the common difference, n is the number of terms in an AP and \(a_{n}\)

\(a_{n} = a + (n-1) \times d\)

Let’s Work Out Example: Find the number of two digit numbers which are multiple of 5. Solution: Given a = 10, d = 5, \(a_{n}\) From the formula of general term, we have: \(a_{n} = a + (n-1).d\) \(95 = 10 + (n-1). 5\) \((n-1) = 17\) \(n = 18\) Example: Find the 20th term for the given AP 3, 5, 7, 9, …… Solution: Given a = 3, d = 5-3 = 2, n = 20 \(a_{n} = a + (n-1).d \) \(a_{n} = 3 + (20-1).2 \) \(a_{n} = 3 + 38 \) |

## Sum of First Terms of an AP:

Consider an AP consisting n terms.

First Term = a

Common Difference = d

nth term = \(a_{n}\)

Sum of n term is given as \(S = \frac{n}{2} \left [ 2a + (n-1).d \right ]\)

Proof:

Consider an AP : a, a+d, a+2d, …………., a+(n-1).d

Sum of first n terms = a + (a+d) + (a+2d) + ………. + [a+ (n-1).d] ——————-(i)

Writing the terms in reverse order, we have: S = [a+(n-1).d] + [a+(n-2).d] + [a+(n-3).d] + ……. (a) ———–(ii)

Adding both the equations term wise, we have

2S = [2a + (n-1).d] + [2a + (n-1).d] + [2a + (n-1).d] + …………. + [2a + (n-1).d] (n-terms)

2S = n. [2a + (n-1).d]

\(S = \frac{n}{2} [2a + (n-1).d]\)

Example: Find the sum of first 30 multiples of 4. Solution: Given, a = 4, n = 30, d = 4 We know, \(S = \frac{n}{2} [2a + (n-1).d]\) \(S = \frac{30}{2} [2(4) + (30-1).4]\) \(S = 15 [8 + 116]\) S = 1860 |

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