HOW TO FIND NTH TERM OF AN ARITHMETIC SEQUENCE

The formula to find nth term of an arithmetic sequence :

an = a1 + (n - 1)d

an ----> nth term

a1 ----> 1st term

d ----> common difference 

Examples 1-4 : Find the nth term of the following arithmetic sequences :

Example 1 :

4, 9, 14, …………

Solution :

Common difference :

d = a2 – a1

= 9 – 4

= 5

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 4 and d = 5.

an = 4 + (n - 1)(5)

= 4 + 5n - 5

= 5n - 1

Example 2 :

125, 120, 115, 110, …………

Solution :

Common difference :

d = a2 – a1

= 120 – 125

-5

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 125 and d = -5.

an = 125 + (n - 1)(-5)

= 125 - 5n + 5

= 130 - 5n

Example 3 :

24, 23¼, 22½, 21¾, …………

Solution :

Common difference :

d = a2 – a1

23¼ - 24

= 93/4 - 24

= (93 - 96)/4

= -3/4

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 24 and d = -3/4.

an = 24 + (n - 1)(-3/4)

= 24 - 3n/4 + 3/4

= 24 + 3/4 - 3n/4

= (96 + 3)/4 - 3n/4

= 99/4 - 3n/4

Example 4 :

√2, 3√2, 5√2, …………

Solution :

Common difference :

d = a2 – a1

3√2 - √2

= 2√2

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = √2 and d = 2√2.

an = √2 + (n - 1)(2√2)

√2 + (2√2)n - 2√2

= (2√2)n - √2

Example 5 :

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

Solution :

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 6 :

Find n so that the nth terms of the following two arithmetic sequences are the same.

1, 7, 13, 19, …………

100, 95, 90, …………

Solution :

Part (i) :

1, 7, 13, 19, …………

Common difference :

d = a2 – a1

= 7 - 1

= 6

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 1 and d = 6.

an = 1 + (n - 1)(6)

= 1 + 6n - 6

= 6n - 5

Part (ii) :

100, 95, 90, …………

Common difference :

d = a2 – a1

= 95 - 100

= -5

nth term of an arithmetic sequence :

an = a1 + (n - 1)d

Substitute a1 = 100 and d = -5.

an = 100 + (n - 1)(-5)

= 100 - 5n + 5

= 105 - 5n

Solve for n :

Given : nth terms of the two arithmetic sequences are the same.

6n - 5 = 105 - 5n

11n - 5 = 105

11n = 110

n = 10

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