EXAMPLE PROBLEMS IN ARITHMETIC SEQUENCE

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

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

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

Substitute 6 for a₁ and 5 for d. 

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

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

General Term : 

a_n  =  a₁ + (n - 1)d

Substitute 6 for a₁ 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 :

a₁  =  125

a₂  =  120

Common Difference :

d  =  a₂ - a₁  

d  =  120 - 125

  d  =  -5

15^th Term :

a_n  =  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 - a₁  =  23¼ - 24  =  -3/4

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

Then, 

a_n  =  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 29^th 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 10^th and 18^th terms of an arithmetic sequence are 41 and 73 respectively. Find the 27th term

Solution :

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

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