SEQUENCES AND SERIES
Definition: A sequence is a set of numbers in a defined order with a rule for obtaining each
of the numbers. The elements of a sequence are called terms of the sequence where the nth
term is denoted by un .
Example:
1. 5, 10, 15, 20, · · · (positive integer multiples of 5 with general term 5r, r is a positive
integer)
2. 1, −3, 9, −27, · · · (powers of −3 with general term (−3)k , k = 0, 1, 2, 3, · · · , i.e. k is a
non-negative integer)
3. 27, 64, 125, 216, · · · (cubes of consecutive integers in ascending order starting with 33 ,
having general term a3 , a = 3, 4, 5, 6, · · · )
Definition: A sequence that is made up of a finite number of terms is called a finite sequence;
otherwise, it is callled an infinite sequence.
Example:
1. The sequence 10, 20, 30, · · · , 100 is finite (with 10 terms).
2. The sequence 10, 20, 30, · · · is infinite (has infinitely many terms).
Definition: A series is obtained by adding the terms of a sequence. A finite series is one
that corresponds to a finite sequence while an infinite series is one that corresponds to an
infinite sequence.
Example:
10 10
10k and 33 + 43 + 53 + · · · + 103 = r3 are finite series.
P P
1. 10 + 20 + 30 + · · · + 100 =
k=1 r=3
∞
P k
2. 1 + 1/4 + 1/16 + 1/64 + 1/256 + · · · = (1/4) is an infinite series (note that k has no upper
k=0
limit in this case).
Arithmetic Progression (AP)
Definition: An arithmetic progression (AP) is a sequence in which successive terms differ
by a constant number. In this case, the constant number is called the common difference
(cd). If an AP has its first term a and its common differnce d, then its nth term is given by
un = a + (n − 1)d.
Example:
1. Determine the first six terms of the AP whose first term is 18 and common difference
is −5.
Solution: In this case a = 18, d = −5. Thus,
u1 = a = 18, u2 = a + d = 18 + (−5) = 13, u3 = a + 2d = 18 + 2(−5) = 8,
u4 = a + 3d = 18 + 3(−5) = 3, u5 = a + 4d = 18 + 4(−5) = −2,
u6 = a + 5d = 18 + 5(−5) = −7.
1
, 2. Determine the 4th , 12th and nth terms of the AP whose first term is 19 and common
difference is 6.
Solution: In this case a = 19, d = 6. Hence,
u4 = a + 3d = 19 + 3(6) = 37, u12 = a + 11d = 19 + 11(6) = 85,
un = a + (n − 1)d = 19 + 6(n − 1) = 13 + 6n.
The Sum of an AP
To find the sum of any AP whose first term, cd, and the number of terms to be added are
all known, the method of first principles can be used.
Example:
Add up the integers from 1 to 100.
Solution:
1 + 2 + 3 + · · · + 98 + 99 + 100
100 + 99 + 98 + · · · + 3 + 2 + 1
101 + 101 + 101 + ··· + 101 + 101 + 101 = 101 × 100
= 10100
10100
Therefore, 1 + 2 + 3 + · · · + 98 + 99 + 100 = 2
= 5050.
A general AP whose first term is a, whose common difference is d, and with n terms is given
as a, a + d, a + 2d, a + 3d, · · · , a + (n − 2)d, a + (n − 1)d. The method used in the example
above can be used to determine a formula for the sum Sn of the n terms of the AP as follows:
Sn = a + [a + d] + [a + 2d] + ··· + [a + (n − 2)d] + [a + (n − 1)d]
Sn = [a + (n − 1)d] + [a + (n − 2)d] + [a + (n − 3)d] + ··· + [a + d] + a
2Sn = [2a + (n − 1)d] + [2a + (n − 1)d] + [2a + (n − 1)d] + ··· + [2a + (n − 1)d] + [2a + (n − 1)d] = n [2a + (n − 1)d
Therefore, Sn = n2 [2a + (n − 1)d].
Example:
1. The fourth term of an AP is 13 and the seventh term is 22. Determine:
(a) the common difference.
(b) the value of n if the nth term is 100.
(c) the value of m if the sum to m terms of the AP is 175.
Solution:
a + 3d = 14
(a) In this case we have . Solving simultaneously gives a = 4, d = 3 so
a + 6d = 22
that cd = 3.
(b) Now, un = 4 + 3(n − 1) = 100. So, n = 33.
(c) We have Sm = m2 [2 (4) + 3 (m − 1)] = 175 so that m [8 + 3m − 3] = 350. On
rewriting this becomes 3m2 + 5m − 350 = 0. Factoring the quadratic function
we get (3m + 35) (m − 10) = 0 so that m = −35/3 or m = 10. Since m must be a
2
Definition: A sequence is a set of numbers in a defined order with a rule for obtaining each
of the numbers. The elements of a sequence are called terms of the sequence where the nth
term is denoted by un .
Example:
1. 5, 10, 15, 20, · · · (positive integer multiples of 5 with general term 5r, r is a positive
integer)
2. 1, −3, 9, −27, · · · (powers of −3 with general term (−3)k , k = 0, 1, 2, 3, · · · , i.e. k is a
non-negative integer)
3. 27, 64, 125, 216, · · · (cubes of consecutive integers in ascending order starting with 33 ,
having general term a3 , a = 3, 4, 5, 6, · · · )
Definition: A sequence that is made up of a finite number of terms is called a finite sequence;
otherwise, it is callled an infinite sequence.
Example:
1. The sequence 10, 20, 30, · · · , 100 is finite (with 10 terms).
2. The sequence 10, 20, 30, · · · is infinite (has infinitely many terms).
Definition: A series is obtained by adding the terms of a sequence. A finite series is one
that corresponds to a finite sequence while an infinite series is one that corresponds to an
infinite sequence.
Example:
10 10
10k and 33 + 43 + 53 + · · · + 103 = r3 are finite series.
P P
1. 10 + 20 + 30 + · · · + 100 =
k=1 r=3
∞
P k
2. 1 + 1/4 + 1/16 + 1/64 + 1/256 + · · · = (1/4) is an infinite series (note that k has no upper
k=0
limit in this case).
Arithmetic Progression (AP)
Definition: An arithmetic progression (AP) is a sequence in which successive terms differ
by a constant number. In this case, the constant number is called the common difference
(cd). If an AP has its first term a and its common differnce d, then its nth term is given by
un = a + (n − 1)d.
Example:
1. Determine the first six terms of the AP whose first term is 18 and common difference
is −5.
Solution: In this case a = 18, d = −5. Thus,
u1 = a = 18, u2 = a + d = 18 + (−5) = 13, u3 = a + 2d = 18 + 2(−5) = 8,
u4 = a + 3d = 18 + 3(−5) = 3, u5 = a + 4d = 18 + 4(−5) = −2,
u6 = a + 5d = 18 + 5(−5) = −7.
1
, 2. Determine the 4th , 12th and nth terms of the AP whose first term is 19 and common
difference is 6.
Solution: In this case a = 19, d = 6. Hence,
u4 = a + 3d = 19 + 3(6) = 37, u12 = a + 11d = 19 + 11(6) = 85,
un = a + (n − 1)d = 19 + 6(n − 1) = 13 + 6n.
The Sum of an AP
To find the sum of any AP whose first term, cd, and the number of terms to be added are
all known, the method of first principles can be used.
Example:
Add up the integers from 1 to 100.
Solution:
1 + 2 + 3 + · · · + 98 + 99 + 100
100 + 99 + 98 + · · · + 3 + 2 + 1
101 + 101 + 101 + ··· + 101 + 101 + 101 = 101 × 100
= 10100
10100
Therefore, 1 + 2 + 3 + · · · + 98 + 99 + 100 = 2
= 5050.
A general AP whose first term is a, whose common difference is d, and with n terms is given
as a, a + d, a + 2d, a + 3d, · · · , a + (n − 2)d, a + (n − 1)d. The method used in the example
above can be used to determine a formula for the sum Sn of the n terms of the AP as follows:
Sn = a + [a + d] + [a + 2d] + ··· + [a + (n − 2)d] + [a + (n − 1)d]
Sn = [a + (n − 1)d] + [a + (n − 2)d] + [a + (n − 3)d] + ··· + [a + d] + a
2Sn = [2a + (n − 1)d] + [2a + (n − 1)d] + [2a + (n − 1)d] + ··· + [2a + (n − 1)d] + [2a + (n − 1)d] = n [2a + (n − 1)d
Therefore, Sn = n2 [2a + (n − 1)d].
Example:
1. The fourth term of an AP is 13 and the seventh term is 22. Determine:
(a) the common difference.
(b) the value of n if the nth term is 100.
(c) the value of m if the sum to m terms of the AP is 175.
Solution:
a + 3d = 14
(a) In this case we have . Solving simultaneously gives a = 4, d = 3 so
a + 6d = 22
that cd = 3.
(b) Now, un = 4 + 3(n − 1) = 100. So, n = 33.
(c) We have Sm = m2 [2 (4) + 3 (m − 1)] = 175 so that m [8 + 3m − 3] = 350. On
rewriting this becomes 3m2 + 5m − 350 = 0. Factoring the quadratic function
we get (3m + 35) (m − 10) = 0 so that m = −35/3 or m = 10. Since m must be a
2
