# EXAMPLE PROBLEMS IN ARITHMETIC SEQUENCE

In this section, we are going to see some example problems in arithmetic sequence.

General term or nth term of an arithmetic sequence :

an  =  a1 + (n - 1)d

where 'a1' is the first term and 'd' is the common difference.

Formula to find the common difference :

d  =  a2 - a1

Formula to find number of terms in an arithmetic sequence :

n  =  [(l - a1) / d] + 1

where 'l' is the last term, 'a1' is the first term and 'd' is the common difference.

General form of an arithmetic sequence :

a1, (a+ d), (a+ 2d), (a+ 3d),.........

where 'a1' is the first term and 'd' is the common difference.

## Arithmetic Sequence - Example Problems

Problem 1 :

The first term of an A.P is 6 and the common difference is 5. Find the arithmetic sequence its general term.

Solution :

a1  =  6

d  =  5

Arithmetic Sequence :

a1, (a+ d), (a+ 2d), (a+ 3d),....................

Substitute 6 for a1 and 5 for d.

6, (6 + 5), (6 + 2 ⋅ 5), (6 + 3 ⋅ 5),....................

6, 11, 16, 21,....................

General Term :

an  =  a1 + (n - 1)d

Substitute 6 for a1 and 5 for d.

an  =  6 + (n - 1)5

an  =  6 + 5n - 5

an  =  5n + 1

Problem 2 :

Find the common difference and 15th term of an arithmetic sequence :

125, 120, 115, 110,...............

Solution :

a1  =  125

a2  =  120

Common Difference :

d  =  a2 - a1

d  =  120 - 125

d  =  -5

15th Term :

an  =  a1 + (n - 1)d

Substitute 15 for n, 125 for a1 and -5 for d.

a15  =  125 + (15 - 1)(-5)

a15  =  125 + (14)(-5)

a15  =  125 - 70

a15  =  55

Problem 3 :

Which term of the arithmetic sequence 24, 23¼, 22½, 21¾, ……… is 3?

Solution :

a1  =  24

d  =  a2 - a=  23¼ - 24  =  -3/4

Let 3 be the nth term of the given arithmetic sequence.

Then,

an  =  3

a1 + (n - 1)d  =  3

Substitute 24 for a1 and -3/4 for d.

24 + (n - 1)(-3/4)  =  3

96/4 - 3n/4 + 3/4  =  3

(96 - 3n + 3) / 4  =  3

(-3n + 99) / 4  =  3

Multiply each side by 4.

-3n + 99  =  12

Subtract 99 from each side.

-3n  =  -87

Divide each side by -3.

n  =  29

Therefore, 3 is 29th term in the given arithmetic sequence.

Problem 4 :

How many terms are in the following arithmetic sequence ?

7, 11, 15,……………483

Solution :

a1  =  7

d  =  a2 - a1  =  11 - 7  =  4

Formula to find number of terms in an arithmetic sequence :

n  =  [(l - a1) / d] + 1

Substitute 483 for l, 7 for a1 and 4 for d.

n  =  [(483 - 7) / 4] + 1

n  =  [476 / 4] + 1

n  =  119 + 1

n  =  120

Therefore, there are 120 terms in the given arithmetic sequence.

Problem 5 :

The 10th and 18th terms of an arithmetic sequence are 41 and 73 respectively. Find the 27th term

Solution :

 a10  =  41a1 + (10 - 1)d  =  41a1 + 9d  =  41 -----(1) a18  =  73a1 + (18 - 1)d  =  73a1 + 17d  =  73 -----(2)

Solving (1) and (2),

a1  =  5

d  =  4

27th Term :

a27  =  a1 + (27 - 1)d

a27  =  a1 + 26d

Substitute 5 for a1 and 4 for d.

a27  =  5 + 26(4)

a27  =  5 + 104

a27  =  109 After having gone through the stuff given above, we hope that the students would have understood how to solve problems in arithmetic sequence.

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