|
A sequence is a list of numbers
arranged in an order that depends upon the position of each term in
the list.
For example,
sequence 1: 3, 7, 11, 15, 19, 23
is a sequence. The numbers are called
terms of the sequence.
Notation:
|
3
|
7
|
11
|
15
|
19
|
23
|
|
t1
|
t2
|
t3
|
t4
|
t5
|
t6
|
t1 is
the first term
and its value is 3
t2
is the second term and its value s 7
t3
is the third term and its value is 11
t4
is the fourth term and its value is 15
t5
is the fifth term and its value us 19
t6
is the sixth term and its value is 23
Pay close attention to the
relationship of the subscript used with t to the term it represents.
t5
<--> 5th
term and so forth. Until this relationship is firmly mastered, it
helps to write the terms t1,
t2, etc., above or
below the actual values in the sequence to
answer
questions about the sequence.
Sequence 1 is
called a finite sequence. It starts with three and ends
with 23.
We can turn this
sequence into one with no last term by added an ellipsis at the end
like so:
3, 7, 11, 15,
19, 23, ...
Now this sequence
is called an infinite sequence, i.e., not finite.
The point of
using the letter t with a subscript is to allow us to use formulas to
specify the relationship between one term and the next. We can talk
about the sequence then in general terms without having to specify each
specific term. This relationship between terms specifies the type
of sequence we have. In particular, if a constant difference exists
between terms as in this sequence then we have an arithmetic
sequence, if a constant ratio exists between terms then we have a
geometric sequence. These two types will be our focus for now.
(Other sequences exist.)
Let's look at this
sequence again.
|
3
|
7
|
11
|
15
|
19
|
23
|
|
t1
|
t2
|
t3
|
t4
|
t5
|
t6
|
|
|
3 + 4
|
7 + 4
|
11 + 4
|
15 + 4
|
19 + 4
|
Each term is 4 more than the previous term, and since this is a constant
difference, this sequence is an arithmetic sequence.
If we use the letter d for difference, then d
= 4. (note, d does not need to be a member of the Natural Numbers
N
.)
Let's have a another look.
|
3
|
7
|
11
|
15
|
19
|
23
|
|
t1
|
t2
|
t3
|
t4
|
t5
|
t6
|
|
|
3 + 4
|
3 + 4 + 4
|
3 + 4 + 4 + 4
|
3 + 4 + 4 + 4 + 4
|
3 + 4 + 4 + 4 + 4 +4
|
|
|
3 + 1(4)
|
3 + 2(4)
|
3 + 3(4)
|
3 + 4(4)
|
3 + 5(4)
|
|
|
t1 + d
|
t1 + 2d
|
t1 + 3d
|
t1 + 4d
|
t1 + 5d
|
The third row shows how many times we
need to add four to the first term to get every other term. The
second term we add once, the third term we add twice, etc., until the
6th term we add 5 times. The last row shows this relationship. In
each case notice that the number of d's we add is one less then the
subscript of the term itself.
For example t5
is: t1 + (5 - 1)d =
t1 + 4d
We use this relationship to find the
value of any term. For example, the value of the 50th
term:
t50
= t1 + (50 -
1)4 = 3 + (49)4 = 3 + 196 = 199
If we let n take on the values 1, 2,
3, ... then we can in general write
for integer n > 0,
|
eqn 1:
|
tn
= t1
+ ( n - 1)•d
|
use to find any term using the first term
|
|
eqn
2:
|
tn = tn-1 + d
|
use to find terms one after the other
|
remember,
the SUBSCRIPT n is the same n in the equation:
tn
= t1 + ( n - 1)
•
d
tn
= tn-1
+ d
Comment
1:
Eqn 2 is the the relationship used in inductive
proofs. It is called a recursive definition.
Comment 2:
In general we can think
of eqn 1 as a function that takes the Natural Numbers
N into the values of the sequence which belong to
the Real Numbers R
. So we have a mapping from
N to R
. (The sequence can be generalized to contain
complex numbers, or any other mathematical object.)
Examples:
|
1) What is the 23rd
term in sequence 1?
t23 =
t1 + (23 - 1) • 4
t23 = 3
+ (22) • 4
t23
= 3 + 88 = 91
|
direct
substitution into eqn 1
|
|
2) Is
55 a member of sequence 1? If so which term is it?
tn
= 3 + (n - 1) •
4
55 = 3 + 4n - 4
56 = 4n
14 = n
yes,
it is the 14th
term
|
n
must be a member of
N to be a
solution. If it is fractional then tn
is not a term in this sequence.
|
|
3) Is
10,132 a member of sequence 1? If so which member?
tn
= 3 + (n - 1) •
4
10,132 = 3 + 4n
- 4
10,133 = 4n
2533.25 = n
no,
it is not a member
|
n is
fractional, so 10,132 is not a member of the sequence
|
|
4)
What is the 5th term of the following sequence? the 61st
term?
the
62nd term? the 60th term?
1,
6, 11, ...
Looking
for a pattern we see that
1 + 5
= 6 and 6 + 11 = 5, which means we have an arithmetic sequence
with difference d = 5
1,
6, 11, 16, 21 t5 = 21
t61
= 1 + (61 - 1) •
5
t61
= 1 + 60 •
5
t61
= 301
t62 =
t61 + d
t62
= 301 + 5
t62
= 306
t61
= t60 + d
t61
− d = t60
301
− 5 = t60
t60
= 296
|
This
can be done without eqn 1 to
find
the 5th term, by writing out the sequence.
Eqn 1
must be used here to find the 61st term.
Direct
substitution into eqn 2 answers these questions, as would
direct
addition or subtraction.
|
|
5)
Given the 41st term of an arithmetic sequence is 125 and
d = 3,
what is the first term?
tn
= t1 + (n - 1)d
125
= t1 + (41 - 1) 3
125
= t1 + (40) 3
125
- 120 = t1
t1
= 5
|
Direct
substitution into eqn 1 solves this problem.
|
|
6)
Suppose the 21st term of an arithmetic sequence is 121 and the
first term is 1, what is the difference d?
121
= 1 + (21 - 1 ) d
120
= 20 d
d
= 6
|
Direct
substitution into eqn 1 solves this problem.
the
sequence is:
1,
7, 13, 19, ...
|
|
7)
Find d in the sequence
2,
2½, 3, 3½, 4, 4½, ...
2½
- 2 = ½
d
= ½
|
|
|
8)
The 3rd term in an arithmetic sequence is 8 and the 16th
is 47. Write the first 5 terms of this sequence
Consider:
|
t3
|
t4
|
t5
|
t6
|
t7
|
t8
|
t9
|
t10
|
t11
|
t12
|
t13
|
t14
|
t15
|
t16
|
|
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
+d
|
|
The
table shows that we add d to t3
, 13 times, to
reach t16.
That
is
t16
= t3 +
13d
47
= 8 + 13d
39
= 13d
d
= 3
t3
= t1 + (3 - 1 )d
t3
= t1 + 2d
t3
- 2d
= t1
t1
= t3 - 2d
t1
= 8 - 2(3)
t1
= 8 - 6 = 2
The
sequence is: 2, 5, 8, 11, 14 ...
We
could have simply subtracted 3 from 8 twice to get 2,
(the
next to last step above shows this to be true)
|
note
(also):
16
- 3 = 13
|
|
9) Write the
terms of the arithmetic sequence given by:
tn
= 2+ 4(n - 1)
t1
= 2 and d = 4, so immediately
2, 6, 10, 14,
...
|
Although this
may not look like an arithmetic sequence, you can see it is by
writing the first three terms:
2 + 4(1-1) = 2
2 + 4(2-1) = 6
2 + 4(3-1) = 10
d = 4
In general, we
consider the difference between
tn+1
and
tn
|
|
In
mathematics we are not usually interested in specific terms or
results, we need to generalize the property then prove the property
always under certain conditions. In this case the condition is, we
have an arithmetic sequence.
In the sequence of problem
9, we see that the difference d between
adjacent term is 4. So
let's assume that all the way up to the nth term d is 4.
Let's show that the
(n+1) term is 4 more than the nth
term.
Substituting
into equation one for each term and subtracting we get:
d =
2 + 4((n+1)
- 1) - (2 + 4(n - 1))
d = 2 + 4(n) -
2 - (4n - 4)
d = 2 + 4n - 2
- 4n + 4
d = 2 - 2 +
4n - 4n + 4
so d = 4
for every pair of adjacent terms.
*advanced*:
I mentioned
previously that sequences form the basis of a technique of
mathematical proof called proof by induction. In
essence we have just done the work to prove that the difference
between any two adjacent terms in this sequence is always 4. The
way this type of proof proceeds is:
1) Show
that the difference between the first two terms is 4.
2) Assume
that the difference between the nth and (n-1) terms is
4.
3) Show
that the difference is 4 between the (n + 1) and nth terms (that,
is, the next term from tn).
We just did
this,
see previous example.
So we have
proven that the difference between any two terms of this sequence
is d.
It is
customary to write at the end of the proof the three letters:
QED
which comes
from Latin, quod erat demonstrandum,
which means
that
which was to be demonstrated.
ED signifies
the end of the proof.
When
we demonstrate a property, we show its correctness in a given
case. Demonstration is not a proof. The must be a strong
connection between the general case to the specific case
before the demonstration can be called a proof, i.e., the
property must hold in general. In mathematics we are not that
interested in specific cases, but in the general case proven to hold
true.
|
|
10)
Write the terms of the arithmetic sequence given by:
tn
= -5 + (1.2)(n - 1)
t1 =
-5 and d = 1.2, so
directly (with a calculator if needed)
-5
+ -3.8 + -2.6 + -1.4 + -0.2 + 1 + 2.2 + ...
|
|
Consider the following
sequence:
3, 9, 27, 54, 162,
486, 1458, 4347, ...
If we find the
difference between any two adjacent terms we find that the difference
is not constant: 9 - 3 = 6, 27 - 9 = 18, 54 - 27 =
27, etc.
This sequence is
therefore not arithmetic, if a constant ratio exists between each successive
pair of terms, the this sequence is a geometric sequence.
Consider the ratios
between adjacent terms:
9/3 = 3, 27/9 =
3, 54/27 = 3, 162/54 = 3, 486/162 = 3,
1458/486 = 3
This ratio is always 3.
We use the letter r to stand for ratio and in this case
r = 3.
|
sequence
|
3
|
9
|
27
|
54
|
162
|
486
|
1458
|
4347
|
|
term
|
t1
|
t2
|
t3
|
t4
|
t5
|
t6
|
t7
|
t8
|
|
n
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
|
products
|
3•1
|
3•3
|
3•3•3
|
3•3•3•3
|
3•3•3•3•3
|
3•3•3•3•3•3
|
3•3•3•3•3•3•3
|
3•3•3•3•3•3•3•3
|
|
r=3
|
3•30
|
3•31
|
3•32
|
3•33
|
3•34
|
3•35
|
3•36
|
3•37
|
Each term is the first
term multiplied by 3 raised to the power n - 1. This sequence is
written
t n
= 3 •
3 n - 1
The
general equation of a geometric sequence is:
tn
= t1
• r
n-1
(
for integer n > 0 )
Examples:
|
1)
write the first 5 terms of the sequence tn = 2
•
3n-1
t1
= 2 •
31-1 = 2
•
30 = 2
t2
= 2 •
32-1 = 2
•
31 = 6
t3
= 2 •
33-1 = 2
•
32 = 18
t4
= 2 •
34-1 = 2
•
33 = 54
t5
= 2 •
35-1 = 2
•
34 = 162
we
have: 2, 6, 18, 54, 162
|
by direct substitution
|
|
2) 3,
6, 12, 24, 48
find
r: 6/3 = 2, 24/12 = 2 so
r = 2.
so tn
= 3•
2n-1
|
Only
the first ratio needs to be calculated, however, another will
reinforce the correctness of that ratio.
|
|
3)
1000, 1060, 1123.60, ...
find r:
1060/1000 =
1.06,
1123.60/1060 =
1.06
r =
1.06
so tn
= 1000 •1.06n-1
|
the
10th term would be:
t10
= 1000(1.069) = 1689.48
the
56th term is:
t56
= 1000(1.0655) = 24,650.32
|
|
4) 1,
-1/3, 1/9, -1/27, ...
find
r:
(-1/3)
/ 1 = -1/3
(1/9)
/(-1/3) = -1/3
so tn
= 1
•
(-1/3) n-1
|
the
10th term would be:
t10
= 1((-1/3)9) =
(-1)9 /
(3)9 = -1/19683
the
7th term would be
t7
= 1((-1/3)6 =
(-1)6
/ (3)6 =
1/3729
|
|
5) if
the 4th term of a geometric sequence is 270 and the
ratio is 3,
what
is the first term?
Substituting
into the general equation
we
have
t5
= t1 (34-1)
270
= t1 (33)
270
= 27 t1
270/27
= t1
10
= t1
so:
t1 = 10
|
|
|
6) if
the 8th term of a sequence is 768 and the 12th
term of a sequence is 12288, write the first 4 terms of this
sequence
We
need to visualize where these terms are since we need to find the
ratio r
then
the first term t1 to answer this question.
|
sequence
|
768
|
|
|
|
12288
|
|
n
|
8
|
19
|
10
|
11
|
12
|
|
|
|
*
r
|
*
r
|
*
r
|
*
r
|
|
In general for a problem such as this one, subtract
the indices of the terms to find the number of times r is multiplied
to get from one to the next; In this case 12
- 8 = 4.
To get
to 12288 from 768 we need to multiply by r, 4 times, i.e., r4
times
So:
768
r4 = 12288
r4
= 12288/768 = 16
r
= 16 1/4 = 2
(Actually, if
complex numbers are allowed for r, there can be 4 distinct ratios
for this sequence, this will be discussed in the next part.
For now we'll use only one, r = 2.)
Using
either term we can find t1 :
t8
= 768 = t1 28-1
= t1
27 = 128 t1
t1
= 768 / 128 = 6
So our
general equation is: tn = 6
•
2n-1
To
answer the question we substitute into this equation, n=1,2,3,4
or,
more directly multiply t1 by 2, 3 times.
the
first 4 terms of this sequence is: 6, 12, 24, 48
|
|
|