Sequences

1. Arithmetic sequences
Contents
1. Linear sequences.
2. Formula for the next term of an
arithmetic sequence.
3. Formula for the sum of an arithmetic
sequence.
4. Questions and problem solving.
Objectives: to be able to,
•find the nth term of an arithmetic
sequence,
•find the sum of an arithmetic sequence.
Prior knowledge: you should already know
how to,
•solve of quadratic equations,
•solve simultaneous equations.

2. Find thenext two terms and aformulafor thenth termof thesequences below.
Arithmetic sequences
1. 5, 8, 11, 14, 17, .......
2.
1
2
, 1,
3
2
, 2,
5
2
, .......
3. 15, 13, 11, 9, 7, .......
20,23 3n2
3,
7
2
1
2
n
5,3 172n
Each sequence is known as linear or arithmetic.
We are adding on a common difference, d, in each sequence.
d  3
d 
1
2
d  2
In theabovewehavegiven theformulaein terms of n. At IB thenotation is slightly different.

3. The nth term of an arithmetic sequence.
Consider thesequence5, 8, 11, 14, 17, ....... We need two numbers to find the nth term
- the difference, d and the first term u1
.
u1
 5, d  3
Write down u2
(the second term) in
terms of u1
and d.
u2
 53u1
d
Write down u3
(the third term) in
terms of u1
and d.
Write down u4
(the fourth term) in
terms of u1
and d.
What about the nth term un
?
u3
 56 u1
2d
u4
 59u1
3d
 
1 1
n
u u n d
  

4. The nth term of an arithmetic sequence.
 
1
Use the general formula 1 to write the th term of each of the following
arithmetic sequences.
n
u u n d n
  
1. 7, 12, 17, 22, 27, .....
2. 6, 8, 10, 12, 14, .....
3. 20, 17, 14, 11, 8, .....
4.
1
4
,
1
2
,
3
4
,1,
5
4
, ......
5. 1,
7
8
,
3
4
,
5
8
,
1
2
......
un
 75(n1)un
 25n
un
 42n
un
 233n
un

1
4
n
un

9
8

1
8
n

5. The Gauss Problem
This is the 18th century German mathematician
Johann Carl Friedrich Gauss. A story is told of how
his lazy mathematics teacher set the class a
problem of adding all the integers from
1 to 100.
His teacher sat back for a rest expecting the class to
spend an hour doing this calculation, but Gauss
gave him the answer in a matter of minutes.
How did he do it?
The answer is 5050. How can you do it without a
calculator in a matter of minutes like Gauss?

6. The Gauss Problem continued
1 + 2 + 3 + 4 + 5 + ………….. 95 + 96 + 97 + 98 + 99 +100
Gauss started to pair off numbers at
the start and end of the sum.
1 + 100=101
2 + 99=101
3 + 98=101
4 + 97=101
How many pairs of these numbers
did Gauss have?
101
50
5050
50 pairs of 101
Gauss’ method can be applied to
any arithmetic sequence.

7. Sum of an arithmetic sequence
Use the Gauss technique to add up the sequence, 1 + 3 + 5 + 7 + …. + 23.
How many terms in this sequence?
un
 u1
d(n1)
12(n1)23
2n2  22
2n  24
n  12
We have 12 terms, or 6 pairs.
The sum of each pair?
1 + 3 + 5 + 7 + …. + 21+ 23
1 + 23=24
3 + 21=24
24
6
144
What is the relationship between the number
of terms, n, and the number of pairs. Number of pairs =
n
2

8. Sum of an arithmetic sequence
Add up the first 9 terms of the sequence 5 + 9 + 13 + …. + 37.
Pair off the numbers 5 + 9 + 13 + 17 + 21 + 25 + 29 + 33 + 37 5 + 37=42
9 + 33=42
13 + 29=42
17 + 25=42
We have 4 pairs of 42 and 21.
or 4
1
2
pairs of 42.
Sum = 4
1
2
 42 = 189
Can we now develop a general formula for the sum of an arithmetic sequence.

9. The general sum of an arithmetic sequence
Sn
 u1
u2
u3
.........un
       
1 1 1 1 1
2 3 ......... ( 1)
n
S u u d u d u d u n d
          
From previous work we have found that
we add together the first and last term and
multiply by the number of pairs -
n
2
.
 
1 1 ( 1)
2
n
n
S u u n d
 
    
 
 
1 1 ( 1)
2
n
n
S u u n d
   
 
1
2 ( 1)
2
n
n
S u n d
  
Thegeneral termfor thesumof an arithmetic sequence.

10. Sum of an arithmetic sequence un
 u1
 n1
 d  
1
2 ( 1)
2
n
n
S u n d
  
1. Find the sum of the first 21 terms of the
sequence 5, 7, 9, 11, 13, ......
2. Calculate the sum of
1
2
1
3
2
2......42
u1
 5 d  2 n  21
 
1
2 ( 1)
2
n
n
S u n d
  
 
21
21
(2)(5) (20)(2)
2
S  
S21
 525
We need the value of .
n un
 u1
 n1
 d
u1

1
2
d 
1
2
un
 42  
1 1
42 1
2 2
n
  
42 
1
2
n
n  84
42
84 1 1
2 (83)
2 2 2
S
 
   
 
   
 
   
 
u1

1
2
d 
1
2
n  84
S42
1785

11. Questions un
 u1
 n1
 d  
1
2 ( 1)
2
n
n
S u n d
  
1. A sequence is defined by un
 253n.
a) Write down the first three terms of the
sequence.
b) Calculatethe30th termof thesequence.
c) Calculate the sum of the first 25 terms of
the sequence.
2. A sequence is defined by un

5n3
2
.
a) Write down the first term u1
and the
common difference, d.
b) Write down the formula for the sum of
the first n terms of this sequence.
c) Calculate down the sum of the first 31
terms of the sequence.
22, 19, 16
65
S25
 350
u1
 4, d 
5
2
 
11 5
11 5
2 2 2 4
n
n n
S n n
 
   
 
 
S31
1286.5

12. Questions  
1 1
n
u u n d
    
1
2 ( 1)
2
n
n
S u n d
  
3. An arithmetic sequence has the first 3
terms k 1, 3k -8, 2k.
Find the value of k and hence find the sum
of the first 25 terms of the sequence.
4. An arithmetic sequence has the first 3
terms 5, k2
-1, 4k -1.
Find two possible solutions for the sum of
the first 10 terms of this sequence.
5. Find the formula for the nth term of an
arithmetic sequence given that u20
 88
and u25
108.
6. By first calculating the number of terms
find the sum of
3
11

7
22

4
11
...
23
11
.
k  5 S25
1000
See solution
k  1 and k  3
S10
 175 and S10
185
See solution
un
 4n8
See solution
n  41 S41

533
11
 48.45
See solution

13. Questions  
1 1
n
u u n d
    
1
2 ( 1)
2
n
n
S u n d
  
7. The sum of the first 3 terms of an arithmetic
sequence is 69 and the sum of the first 5 terms
is 130.
Find the first term and the common difference
of the sequence.
8. An arithmetic sequence is defined as
4, 15, 26, 37, ....
Calculate the first term to exceed 500.
9. The sequence shown below is arithmetic.
5, ..... , ..... , ..... , ..... , 26.
Find the missing numbers in the sequence.
10. Find the smallest value of n such that
the sum of the arithmetic sequence defined
by un

4n1
3
exceeds 100.
u1
 20 d  3 See solution
n  48 is the first term
See solution
5, 9.2, 13.4, 17.6, 21.8, 26
See solution
n 13 is the smallest term.
See solution

14. Worked solutions
An arithmetic sequence has the first 3 terms k 1, 3k -8, 2k.
Find the value of k and hence find the sum of the first 25 terms of the sequence.
k 1, 3k-8, 2k
   
3 8 1
d k k
   
d  2k 7 equation 1
   
2 3 8
d k k
  
d  8k equation 2
2k 7  8k
3k 15
k  5
k 1, 3k-8, 2k and k  5
4, 7, 10, ....
u1
 4 d  3 n  25
   
 
25
25
2 4 (24) 3
2
S  
S25
1000
Return

15. Worked solutions
An arithmetic sequence has the first 3 terms 5, k2
-1, 4k -1.
Find two possible solutions for the sum of the first 10 terms of this sequence.
5, k2
-1, 4k -1
   
2
1 5
d k
  
d  k2
6 equation 1
   
2
4 1 1
d k k
   
d  4k k2 equation 2
k2
6  4k  k2
2k2
 4k 6  0
k2
2k 3  0
(k 3)(k 1) 0
k  1 and k  3
Using k  -1: 5, 0, -5
u1
 5, d  -5, n 10
   
 
10
10
2 5 (9) 5
2
S   
S10
 175
Using k  3: 5, 8, 11
u1
 5, d  3, n 10
   
 
10
10
2 5 (9) 3
2
S  
S10
185 Return

16. Worked solutions
Find the formula for the nth term of an arithmetic sequence given that u20
 88 and u25
108.
u20
 88
u1
19d  88
u25
108
u1
24d 108
equation 1
equation 2
5d  20 eqn 2 - eqn 1
d  4
u1
 4(19) 88
u1
12
u1
12 d  4
un
12 4(n1)
un
 4n8
Return

17. Worked solutions
By first calculating the number of terms find the sum of
3
11

7
22

4
11
...
23
11
.
3
11

7
22

4
11
...
23
11
u1

3
11
d 
7
22

3
11
 d 
76
22
d 
1
22
 
3 1
1
11 22
n
u n
  
 
23 3 1
1
11 11 22
n
  
20
11

n
22

1
22
41
22

n
22
n  41
u1

3
11
d 
1
22
n  41
41
41 3 1
2 (40)
2 11 22
S
 
   
 
   
 
   
 
S41

533
11
 48.45
Return

18. Worked solutions
The sum of the first 3 terms of an arithmetic sequence is 69 and the sum of the first 5 terms is 130.
Find the first term and the common difference of the sequence.
 
1
2 ( 1)
2
n
n
S u n d
  
S3
 69
 
1
3
2 2 69
2
u d
 
u1
d  23 equation 1
S5
130
 
1
5
2 4 130
2
u d
 
u1
2d  26 equation 2
d  3 eqn 2 - eqn 1
u1
 20
u1
 20 d  3
Return

19. Worked solutions
An arithmetic sequence is defined as 4, 15, 26, 37, ...
Calculate the first term to exceed 500.
4, 15, 26, 37
u1
 4 d 11
un
 411(n1)
un
11n7
11n7  500
11n  507
n 
507
11
n  47.09
n  48 is the first term
Return

20. Worked solutions
The sequence shown is arithmetic 5, ..... , ..... , ..... , ..... , 26.
Find the missing numbers in the sequence.
5, ..... , ..... , ..... , ..... , 26
5, 5d, 52d, 53d, 54d, 55d
55d  26
5d  21
d 
21
5
Return
5, 9.2, 13.4, 17.6, 21.8, 26

21. Worked solutions
Find the smallest value of n such that the sum of the arithmetic sequence defined
by un

4n1
3
exceeds 100.
un

4n1
3
u1

4(1)1
3

5
3
u2

4(2)1
3
 3
u3

4(3)1
3

13
3
5
3
3
13
3
... 100
u1

5
3
d 
4
3
5 4
2 ( 1) 100
2 3 3
n
n
 
 
  
 
 
 
 
6 4
200
3 3
n n
 
 
 
 
4n2
6  600
2n2
3  300
n2

297
2
n 12.19
n 13 is the smallest term.
Return