HOW TO FIND THE NTH TERM OF AN ARITHMETIC SEQUENCE

Formula for nth term of an arithmetic sequence :

an  =  a1 + (n - 1)d

a1 ---> first term

d ----> common difference

Formula for common difference :

d = a2 - a1

Example 1 :

Find the nth term of the following arithmetic sequence.

6, 9, 12, …………

Solution :

6, 9, 12, …………

Common difference :

d = a2 – a1

= 9 – 6

= 3

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 6 and d = 3.

an = 6 + (n - 1)(3)

= 6 + 3n - 3

= 3n + 3

Example 2 :

Find the nth term of the following arithmetic sequence.

110, 106, 102, …………

Solution :

110, 106, 102, …………

Common difference :

d = a2 – a1

= 106 – 110

= -4

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 110 and d = -4.

an = 110 + (n - 1)(-4)

= 110 - 4n + 4

= 114 - 4n

Example 3 :

Find the 25th term of the arithmetic sequence whose first term is 5 and common difference is 5.

Solution :

Formula for nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 5, d = 5 and n = 25.

a25 = 5 + (25 - 1)(5)

= 5 + 24(5)

= 5 + 120

= 125

Example 4 :

Find the 16th term of the arithmetic sequence whose first term is 3 and common difference is 3.

Solution :

Formula for nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 8, d = 3 and n = 16.

a16 = 8 + (16 - 1)(3)

= 8 + 15(3)

= 8 + 45

=  53

Example 5 :

Find the 200th term of the arithmetic sequence whose first term is 34 and common difference is 15.

Solution :

Formula for nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 34, d = 15 and n = 200.

a200 = 34 + (200 - 1)(15)

= 34 + 195(15)

= 34 + 2925

= 2959

Example 6 :

Find the 22nd term of the arithmetic sequence whose first term is 5/8 and common difference is 1/8.

Solution :

Formula for nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 5/8, d = 1/8 and n = 22.

a22 = 5/8 + (22 - 1)(1/8)

= 5/8 + 21(1/8)

= 5/8 + 21/8

= (21 + 5)/8

= 26/8

= 13/4

Example 7 :

Find the 39th term of the arithmetic sequence whose first term is 3/2 and common difference is 9/4.

Solution :

Formula for nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 3/2, d = 9/4 and n = 39.

a39 = 3/2 + (39 - 1)(9/4)

= 3/2 + 38(9/4)

= 3/2 + 171/2

= (3 + 171)/2

= 174/2

= 87

Example 8 :

Find the 50th term of the following arithmetic sequence :

0.5, 1, 1.5, 2, …………

Solution :

In the arithmetic sequence 0.5, 1, 1.5, 2,...…

a1 = 0.5

d = 1 - 0.5 = 0.5

Formula for nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 0.5, d = 0.5 and n = 50.

a50 = 0.5 + (50 - 1)(0.5)

= 0.5 + 49(0.5)

= 0.5 + 24.5

= 25

Example 9 :

The 10th and 18th terms of an arithmetic sequence are 41 and 73 respectively. Find the nth term.

10th term = 41

a1 + (10 - 1)d = 41

a1 + 9d = 41 ----(1)

18th term = 73

a1 + (18 - 1)d = 73

a1 + 17d = 73 ----(2)

(2) - (1) :

8d = 32

d = 4

Substitute d = 4 in (1).

a1 + 9(4) = 41

a1 + 36 = 41

a1 = 5

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 5 and d = 4.

an = 5 + (n - 1)(4)

= 5 + 4n - 4

= 4n + 1

Example 10 :

The sum of n terms of an arithmetic sequence is 3n2 + 5n. Find its nth term.

Solution :

Given : The sum of n terms is 3n2 + 5n.

Sn3n2 + 5n

Substitute n = 1.

S1 = 3(1)2 + 5(1)

= 3 + 5

= 8

Substitute n = 2.

S2 = 3(2)2 + 5(2)

= 3(4) + 10

= 22

So, the first term of the given arithmetic sequence is 8.

a1 = 8

Sum of first two terms is 22.

First term + Second term = 22

a1 + a2 = 22

8 + a2 = 22

 a2 = 14 

Common difference :

d = a2 – a1

= 14 – 8

= 6

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 8 and d = 6.

an = 8 + (n - 1)(6)

= 8 + 6n - 6

= 6n + 2

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More