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Sequences and
Series
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Sequences
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Arithmetic sequence/progression
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General term of an arithmetic
sequence
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The sum of the first n terms of an arithmetic sequence |
The sum of the first n natural numbers |
Arithmetic sequences, examples
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Sequences
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Sequence
or progression is an ordered set of numbers, either finite or infinite,
such that each term in a sequence can be indexed meaning, can be written
as an algebraic function of its position in the sequence. |
A
finite sequence has a definite number of terms while an infinite
sequence has an infinite number of terms. |
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Arithmetic sequence/progression
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An arithmetic progression is a sequence of
numbers in which the difference between any two successive numbers is
constant. |
The
difference d
between successive terms is called common
difference. |
Therefore,
an arithmetic sequence can be written as |
a1,
a2, a3, a4,
. . . , an
-1,
an, . . .
or a1,
a1 + d, a1 + 2d,
a1
+ 3d, . . . ,
an
-1,
an, . . .
|
where,
a2
= a1 + d, |
a3 = a1 + 2d, |
a4 = a1 + 3d,
and so on, |
thus,
the formula for the nth
term or the general term of an
arithmetic sequence is |
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The sum of the first n terms of an arithmetic sequence |
When
deriving the formula for the sum of the first
n
terms
of an arithmetic sequence
we use the fact that the sum of
every pair of symmetric terms of the sequence is the same, as is the sum
of the first and
last terms in the sequence the same as, the sum of the second and second
to last terms, and so on.
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Therefore,
by adding the sum of all terms of a sequence to the sum of the same
terms written in the inverse |
order,
that is
Sn
= a1 + a2 + a3 +
. . . + an
-2
+ an
-1 +
an, |
and
Sn
= an
+ an
-1 +
an
-2
+ . . . + a3 +
a2 + a1 |
then, 2Sn
= (a1 + an) + (a2 +
an
-1) + (a3 +
an
-2) +
. . . + (a3 + an
-2) + (a2 +
an
-1) + (a1 +
an) |
since
the right side of the equation represents the sum of n
partial sums each of the same value, |
a1 +
an
or 2a1 + (n -
1) · d |
then,
the sum of the first n
terms of the arithmetic sequence |
|
Sn
= (n/2) · (a1 + an)
or
Sn
= (n/2)
· [2a1 + (n -
1) · d] |
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Let
find the
sum of the arithmetic sequence
3,
6, 9, 12, 15 |
we
add the same sequence in inverse order 15,
12, 9, 6, 3 |
it follows that 2Sn
= (3 +
15) + (6 +
12) + (9 +
9) + (12 +
6) + (15 +
3)
= 5
· 18
= 90, |
thus,
2Sn
= n · (a1 + an)
= 5 · (3 +
15)
or Sn
= (n/2)
· (a1 + an)
= 5/2 ·
18 = 45. |
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The sum of the first n natural numbers
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As
natural numbers represents the arithmetic sequence with the first term
a1
= 1
and
common difference
d
= 1,
we substitute these values into
the formula for
the sum of the first n
terms of the arithmetic sequence,
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|
a1
= 1 and
d
= 1
=> Sn
= (n/2)
· [2a1 + (n -
1) · d]
= (n/2) · [2 · 1 + (n -
1) · 1],
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to
obtain, |
Sn
= n · (n + 1)/2 |
the formula for the sum of the first n
natural numbers. |
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Arithmetic sequences, examples
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Example:
The sum of three
successive terms of an arithmetic sequence is 33 and the product is
1232.
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Find
the greatest term.
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Solution:
Since, (1)
a1 + a2 + a3
= 33,
and
(2)
a1
· a2 · a3
= 1232
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then a1 +
a1 + d +
a1 + 2d
= 33
a1 · (a1 + d) · (a1 + 2d)
= 1232
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3a1 + 3d
= 33,
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a1
= 11 -
d
=>
(2)
(11 -
d)
· (11 -
d + d ) · (11 -
d + 2d )
= 1232 |
d
= 3
=>
a1
= 11 -
d = 11 -
3,
121 -
d 2
= 112, |
a1
= 8.
d 2
= 9,
d
= 3, |
Thus,
the given sequence is 8, 11, 14.
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Example:
If the sum of the
first fifteen terms of an arithmetic sequence is 0 and the first term is
21, find the tenth term of the sequence.
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Solution:
S15
= 0
and
a1
= 21,
Sn
= (n/2)
· (a1 + an)
an
= a1 + (n -
1) · d
|
a10
= ?
S15
= (15/2) · (21 + a15)
= 0
-
21
= 21 + 14 · d |
a15
= -
21, 14 d
= -
42, d
= -
3
|
a10
=
a1 + 9d |
a10
=
21 + 9 · ( -
3)
= -
6, a10
= -
6. |
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Example:
Find the sum of all
natural numbers greater than 30 and smaller than 330 whose last digit is
1.
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Solution:
Given numbers form the sequence 31,
41, 51, . . . , 291, 301,
311, 321.
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Therefore,
a1
= 31,
an
= 321 and
d
= 10 |
as an
= a1 + (n -
1) · d |
then 321
= 31 + (n -
1) · 10
Sn
= (n/2)
· (a1 + an) |
290
= (n -
1) · 10
S30
= 30/2 · (31 +
321) |
n -
1
= 29, n = 30.
S30
= 15 · 352
= 5280. |
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Example:
Find the first term
and number of terms of an arithmetic sequence with an
= 15, Sn
= 64
and
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common
difference
d
= 2.
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Solution: Use
the formulas for an
and Sn
to form a system of two equations in two unknowns,
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(1) an
= a1 + (n -
1) · d |
(2) Sn
= (n/2)
· [2a1 + (n -
1) · d] |
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(1) a1 + (n -
1) · 2
= 15,
=>
a1
= 17 -
2n |
(2) (n/2)
· [2a1 + (n -
1) · 2]
= 64,
=>
n
· (a1+ n -
1)
= 64 |
(1) a1
= 17 -
2n =>
(2)
n2 + a1n -
n
= 64 |
n2 + (17 -
2n) · n -
n
= 64, |
n2
-
16n + 64 = 0, |
n1,2
= 8 ± Ö
64 -
64, n
= 8,
a1
= 17 -
2 · 8
= 1. |
Therefore,
the sequence is 1, 3, 5, 7, 9, 11, 13, 15,
. . .
, where a8
= 15
and S8
= 64. |
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Example:
Which arithmetic sequence has the property that the sum of its first
five terms is 35 and the sum of the second and the sixth term is 20. |
Solution: Use
the formulas for Sn
and an
to form a system of two equations,
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Sn
= (n/2)
· [2a1 + (n -
1) · d]
=>
(1)
(5/2) · [2a1 +
4d]
= 35,
a1 +
2d
= 7 |
an
= a1 + (n -
1) · d,
a2 +
a6
= 20
=>
(2)
(a1 +
d) + (a1 + 5d)
= 20,
2a1 +
6d
= 20 |
(2)
a1 +
3d
= 10 |
(2) -
(1) (a1 + 3d)
-
(a1 + 2d)
= 10
-
7,
d
= 3 |
(2)
a1 +
3d
= 10
=>
a1
= 10
-
3d
= 10
-
3 ·
3
= 1,
a1
= 1. |
Thus,
the sequence is 1, 4, 7, 10, 13, 16, 19,
. . .
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Example:
How many natural numbers, divisible by 7, lie between 0 and 100, and
what is their sum. |
Solution: The
sequence of numbers divisible by 7 is; 7,
14, 21, 28,
. . .
, 84, 91, 98
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and
since an
= a1 + (n -
1) · d,
then 98
= 7 + (n -
1) · 7
| ¸
7 |
Sn
= n/2 · (a1 + an),
n
= 14, |
S14
= 14/2 · (7 +
98)
= 7
· 105
= 735. |
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Example:
Find n
and Sn
of an arithmetic sequence if
given are; a1,
an,
and d. |
Solution: Since,
an
= a1 + (n -
1) · d
then an
-
a1
= (n
-
1) · d,
or
n -
1
= (an
-
a1)
/ d
|
n
= (an
-
a1 +
d) / d
|
and
Sn
= (n/2)
· (a1 + an) |
by
substituting n
= (an
-
a1 +
d) / d
obtained is Sn
= (an
-
a1 +
d) · (a1 + an) / (2d)
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Example:
Find d
and n
of an arithmetic sequence if given are; a1,
an,
and Sn. |
Solution: Since,
an
= a1 + (n -
1) · d
and Sn
= (n/2)
· (a1 + an)
|
d
= (an
-
a1)
/ (n -
1) <=
n
= (2Sn) / (a1 + an) |
d
= (an
- a1)
/ [(2Sn) / (a1 + an) -
1]
= (an2 - a12) / (2Sn
- a1
-
an) |
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Example:
Derive the formula for the nth
odd natural number and the sum of the first n
odd natural numbers. |
Solution: The
sequence of odd natural numbers
is 1, 3, 5, 7,
. . .
, an,
. . .
|
using
an
= a1 + (n -
1) · d |
an
=
1 + (n -
1) · 2
and
since Sn
= (n/2)
· (a1 + an)
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an
= 2n -
1 =>
Sn
then
Sn
= (n/2)
· [1 +
(2n -
1)] so,
Sn
= n2. |
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Example:
In
a sequence of natural numbers divisible by 7 find the one that equals
the one seventeenth of the sum of all natural numbers that precede this
one. |
Solution: A
natural number divisible by 7 is 7n,
where n
Î N,
so the sequence of the numbers is
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7, 14, 21, 28, 35, 42, 49, . . . |
and
since the formula for the sum of the first n
natural numbers Sn
= n · (n + 1)/2, |
and
the sum of natural numbers that precede the number divisible by 7
is, (7n +
1)[(7n -
1) + 1] / 2 |
then,
the following equation meets the given condition |
7n
= (1/17) · 7n(7n -
1) / 2 thus,
7n
= 35. |