# General (nth) Term of AP

A pattern in which each term except the first is obtained by adding a fixed number (positive or negative) to the previous term is called an **Arithmetic Progression (A.P.)**.

The first term is usually denoted by a and the common difference is denoted by d. Thus, standard form of an Arithmetic Progression is:

a, a + d, a + 2d, a + 3d, ...

### General Term

Since first term is a, second term is obtained by adding d to a i.e., a + d, the third term will be obtained by adding d to a + d. So, third term will be (a + d) + d = a + 2d and so on.

First term, t_{1} = a

Second term, t_{2} = a + d

Third term, t_{3} = a + 2d

Fourth term, t_{4} = a + 3d

Each term is a + (term number - 1)d. So,

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

**Example 1: **Find the 15th and nth terms of the AP: 16, 11, 6, 1, -4, -9, ...

Here, a = 16 and d = 11 - 16 = -5

t_{15} = a + (15 - 1)d = a + 14d

= 16 + 14(-5) = 16 0 70

= -54

**Example 2:** The first term of an AP is -3 and 12^{th} term is 41. Determine the common difference.

Let first term of AP be a and common difference be d.

Therefore, t_{12} = a + (12 - 1)d = 41

-3 + 11d = 41

11d = 44

d = 4

**Example 3:** The common difference of an AP is 5 and 10^{th} term is 43. Find its first term.

t_{10} = a + (10 - 1)d

43 = a + 9 × 5

43 = a + 45

a = -2

**Example 4:** The first term of an AP is -2 and 11^{th} term is 18. Find its 15^{th} term.

To find 15^{th} term, you need to find d.

Now t_{11} = a + (11 - 1)d

18 = -2 + 10d

10d = 20

d = 2

Now t_{15} = a + 14d

= -2 + 14 × 2

= 26

**Example 5:** If p times the p^{th} term of an AP is equal to q times the q^{th} term, prove that its (p + q)^{th} term is zero, provided p ≠ q.

t_{p} = a + (p - 1)d

t_{q} = a + (q - 1)d

Since p t_{p} = q tq,

p[a + (p - 1)d] = q[a + (q - 1)d]

pa + p(p - 1)d - qa - q(q - 1)d = 0

(p - q)a + (p^{2} - q^{2})d - pd + qd = 0

(p - q)a + (p^{2} - q^{2})d - (p - q)d = 0

(p - q)a + (p - q)(p + q)d - (p - q)d = 0

(p - q)[a + (p + q)d - d] = 0

a + (p + q)d - d = 0 [as p - q ≠ 0]

a + (p + q - 1)d = 0

Since, LHS (p + q)^{th} term, therefore, t_{p+q} = 0