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

General term or n^{th} term of an arithmetic sequence :

**a _{n} = a_{1} + (n - 1)d**

where 'a_{1}' is the first term and 'd' is the common difference.

Formula to find the common difference :

**d = a _{2} - a_{1}**

Formula to find number of terms in an arithmetic sequence :

**n = [(l - a _{1}) / d] + 1**

where 'l' is the last term, 'a_{1}' is the first term and 'd' is the common difference.

General form of an arithmetic sequence :

**a _{1}, (a_{1 }+ d), (a_{1 }+ 2d), (a_{1 }+ 3d),.........**

where 'a_{1}' is the first term and 'd' is the common difference.

**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 :**

a_{1} = 6

d = 5

**Arithmetic Sequence : **

a_{1}, (a_{1 }+ d), (a_{1 }+ 2d), (a_{1 }+ 3d),....................

Substitute 6 for a_{1} and 5 for d.

6, (6_{ }+ 5), (6_{ }+ 2 ⋅ 5), (6_{ }+ 3 ⋅ 5),....................

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

**General Term : **

a_{n} = a_{1} + (n - 1)d

Substitute 6 for a_{1} and 5 for d.

a_{n} = 6 + (n - 1)5

a_{n} = 6 + 5n - 5

a_{n} = 5n + 1

**Problem 2 : **

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

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

**Solution :**

a_{1} = 125

a_{2}_{ }= 120

**Common Difference :**

d = a_{2} - a_{1 }

d = 120 - 125

d = -5

**15 ^{th} Term :**

a_{n} = a_{1} + (n - 1)d

Substitute 15 for n, 125 for a_{1} and -5 for d.

a_{15} = 125 + (15 - 1)(-5)

a_{15} = 125 + (14)(-5)

a_{15} = 125 - 70

a_{15} = 55

**Problem 3 : **

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

**Solution :**

a_{1} = 24

d = a_{2} - a_{1 }= 23¼ - 24 = -3/4

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

Then,

a_{n} = 3

a_{1} + (n - 1)d = 3

Substitute 24 for a_{1} 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 29^{th} term in the given arithmetic sequence.

**Problem 4 : **

How many terms are in the following arithmetic sequence ?

7, 11, 15,……………483

**Solution :**

a_{1} = 7

d = a_{2} - a_{1 }= 11 - 7 = 4

Formula to find number of terms in an arithmetic sequence :

n = [(l - a_{1}) / d] + 1

Substitute 483 for l, 7 for a_{1} 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 10^{th} and 18^{th} terms of an arithmetic sequence are 41 and 73 respectively. Find the 27th term

**Solution :**

a a a |
a a a |

Solving (1) and (2),

a_{1} = 5

d = 4

**27th Term : **

**a _{27} = a_{1} + (27 - 1)d**

**a _{27} = a_{1} + 26d**

**Substitute 5 for a _{1} and 4 for d. **

**a _{27} = 5 + 26(4)**

**a _{27} = 5 + 104**

**a _{27} = 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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