ARITHMETIC SEQUENCES AND SERIES

Arithmetic Sequence

Example 1 : 

An arithmetic sequence has first term 6 and common difference 8. Find the 12^th term.

Solution :

Formula for n^th term of an arithmetic sequence : 

 a_n  =  a₁ + (n - 1)d

Substitute n = 12, a₁ = 6 and d = 12.

 a₁₂  =  6 + (12 - 1)(8)

=  6 + 11(8)

 =  6 + 88

 =  94

The 12^th term is 94.

Example 2 : 

The third term of an arithmetic sequence is 10 and the 15^th term is 46. Find the first term and the common difference.

Solution :

Use  a_n = a₁ + (n - 1)d to write the given information as system of linear equations. 

Solve (1) and (2). 

(2) - (1) :

12d  =  36

d  =  3

Substitute d = 12 in (1). 

a₁ + 2(3)  =  10

a₁ + 6  =  10

a₁  =  4

The first term is 4 and the common difference is 3.

Example 3 : 

Find the number of terms in the arithmetic sequence :

1, 4, 7,......., 100

Solution :

This is an arithmetic sequence with a₁ = 1 and d = 3. 

Let 100 be the n^th term of this sequence.

a_n  =  100

a₁ + (n - 1)d  =  100

Substitute a₁ = 1 and d = 3.

1 + (n - 1)(3)  =  100

1 + 3n - 3  =  100

3n - 2  =  100

3n  =  102

n  =  34

There are 34 terms in the given arithmetic sequence. 

Arithmetic Series

Example 4 : 

An arithmetic sequence has first term 5 and common difference –2. Find the sum of the first 10 terms.

Solution :

Formula for the sum of first n terms of an arithmetic sequence.

S_n = (n/2)[2a₁ + (n - 1)d]

Substitute n = 10, a₁ = 5 and d = -2.

S₁₀ = (10/2)[2(5) + (n - 1)(-2)]

=  5[10 + (10 - 1)(-2)]

=  5[10 + 9(-2)]

=  5[10 - 18]

=  5(-8)

=  -40

Example 5 : 

An arithmetic sequence has first term 10 and last term 1000. If there are 30 terms, find the sum of all the terms.

Solution :

Formula for the sum of first n terms of an arithmetic sequence.

S_n = (n/2)[a₁ + a_n]

Substitute n = 30, a₁ = 10 and a_n = 1000.

S₃₀ = (30/2)[10 + 1000]

=  15[1010]

=  15150

Example 6 : 

The sum of the first n terms of the sequence 3, 7, 11, 15… is 465. Find n.

Solution :

This is an arithmetic sequence with a₁ = 3 and d = 4.

Sn  =  465

(n/2)[2a₁ + (n - 1)d]  =  465

Substitute a₁ = 3 and 4 = 4.

(n/2)[2(3) + (n - 1)(4)]  =  465

(n/2)(6 + 4n - 4)  =  465

(n/2)(4n + 2)  =  465

n(2n + 1)  =  465

2n² + n  =  465

2n² + n - 465  =  0

Factor and solve. 

2n² + 31n - 30n - 465  =  0

n(2n + 31) - 15(n + 31)  =  0

(n + 31)((n - 15)  =  0

Because n stands for number of terms, it can not be a negative value. 

So, n = 15.

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