Class 10 Maths Arithmetic Progressions | nth term of AP |

**nth term of Arithmetic Progression**

Let a_{1}, a_{2}, a_{3}, . . . be an AP whose first term a1 is a and the common difference is d., thus the AP is Thus a, a + d, a + 2d, a + 3d,

The first term a_{1} = a + 0*d = a + (1 – 1) d

The second term a_{2} = a + d = a + (2 – 1) d

The third term a_{3} = a + 2d = a + (3 – 1) d

The fourth term a_{4} = a + 3d = a + (4 – 1) d

. . . . . . . .

Looking at the pattern, we can say that the nth term a_{n} = a + (n – 1) d.

So, the nth term a_{n} of the AP with first term ‘a’ and common difference d is given by a_{n} = a + (n – 1) d.

a_{n} is also called the general term of the AP

**Example 1: **Find the 10th term of the AP : 2, 7, 12, . . .

**Solution: **Here, *a *= 2, *d *= 7 – 2 = 5 and *n *= 10.

We have *a _{n} *=

So, *a*_{10} = 2 + (10 – 1) × 5 = 2 + 45 = 47

Therefore, the 10th term of the given AP is 47.

**Example 2: **Determine the AP whose 3rd term is 5 and the 7th term is 9.

**Solution: ***a*_{3 }= *a *+ (3 – 1) *d *= *a *+ 2*d *= 5 --(i)

*a*_{7} = *a *+ (7 – 1) *d *= *a *+ 6*d *= 9 --(ii)

Solving the pair of linear equations (i) and (ii), we get a = 3, *d *= 1

Hence, the required AP is 3, 4, 5, 6, 7 . . .

**Example 3 : **Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

**Solution: **d**= ***a*_{2}– *a* = 6 & a=5

Let 301 be a term, say, the *n*th term of this AP.

*a _{n} *=

or 301 = 5 + (*n *– 1) × 6

301 = 6*n *– 1

*Or n = *302/6 which is a fraction. Thus 301 is not in the series.

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