# Arithmetic Progression – Nth Term, AP Formula, Sum of AP & Example

In mathematics, there are different types of progression. However, in this article, I will be discussing the Arithmetic progression (AP). AP is a mathematical sequence in which the difference between two consecutive terms is always a constant.

**Definition**

It is a sequence of numbers such that the difference between any two successive members is a constant. The constant value in this progression is referred to as the common difference.

Mathematically, in arithmetic progression, a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = d. where d is a common difference.

For example, 2, 4, 6, 8, 10, 12, 14 is an example of A.P. The difference between the second term and the first term is 2, and the same is true for third term and second term, etc.

**Notation**

n is the number of terms [for example 1, 4, 7 has 3 terms where (1) is the first term (4) is the second, and (7) is the third term of the sequence]

a_{1} is the first term

T_{n} is the n-th term

d is the common difference

S_{n} is the sum of n number of terms in arithmetic progression

The n-th term of AP is written as, T_{n} = a + (n-1)d

The sum of n number of term of AP is written as, S_{n} = n/2 [2a + (n-1)]d

Read: 8 laws of indices

**Question and Answer**

The second and the sixth terms of an A.P are 10 and 34. Find the (a) common difference (b) first term (c) nth term (d) sum of first 100 terms

Solution

(a)T_{2} = 10, T_{6} = 34

T_{n} = a + (n-1)d

T_{2} = a + (2-1)d = 10 ……. eq 1

T_{6} = a + (6 -1)d= 34 ……… eq 2

A + d = 10 …… eq 1

A + 5d = 34 ….. eq 2

Subtract eq 1 from eq (2)

5d -d = 34 -10

4d = 24, d = 24/4 = 6. So the common difference is 6

(b) From eq 1, a + d = 10, a + 6 = 10, a = 10 – 6 = 4. So the first term is 4

(c) Tn = a + (n-1)d = 4 + (n-1)6 = 4 + 6n – 6 = 6n – 2

So, n-th term is 6n – 2

(d) Sum of first 100 terms

S_{n} = n/2 [2a + (n-1)]d = 100/2 [2*4 + (100-1)*6] = 50(8 + 99*6) = 50(8 + 594) = 50(602) = 30100