Download Arithmetic Sequences . ppt

ARITHMETIC
SEQUENCES
These are sequences where the difference
between successive terms of a sequence
is always the same number. This number
is called the common difference.
3, 7, 11, 15, 19 …
d=4
a=3
Notice in this sequence that if we find the difference
between any term and the term before it we always get 4.
4 is then called the common difference and is denoted
with the letter d.
To get to the next term in the sequence we would add 4
so a recursive formula for this sequence is:
an  an1  4
The first term in the sequence would be a1 which is
sometimes just written as a.
+4 +4 +4 +4
3, 7, 11, 15, 19 …
d=4
a=3
Each time you want another term in the sequence you’d
add d. This would mean the second term was the first
term plus d. The third term is the first term plus d plus d
(added twice). The fourth term is the first term plus d plus
d plus d (added three times). So you can see to get the
nth term we’d take the first term and add d (n - 1) times.
an  a  n  1d
Try this to get the 5th term.
a5  3  5  14  3  16  19
Let’s look at a formula for an arithmetic sequence and see
what it tells us.
We can think of this as
you can see what the
a “compensating term”.
common difference
Without it the sequence
will be in the formula
would start at 4 but this
gets it started where we
want it.
Subbing in the set of positive integers we get:
What is the
common
d=4
difference?
4n 1
3, 7, 11, 15, 19, …
4n would generate the multiples of 4.
With the - 1 on the end, everything is back
one. What would you do if you wanted
the sequence 2, 6, 10, 14, 18, . . .?
4n  2
Find the nth term of the arithmetic sequence when a = 6 and d = -2
If we use -2n we will generate a sequence whose
common difference is -2, but this sequence starts at -2
(put 1 in for n to get first term to see this). We want
ours to start at 6. We then need the “compensating
term”. If we are at -2 but want 6, we’d need to add 8.
 2n  8
Check it out by putting in the first few positive
integers and verifying that it generates our sequence.
6, 4, 2, 0, -2, . . .
Sure enough---it starts at 6 and
has a common difference of -2
Let’s try something a little trickier. What if we just
know a couple of terms and they aren’t consecutive?
The fourth term is 3 and the 20th term is 35. Find the first
term and both a term generating formula and a recursive
formula for this sequence.
How many differences would
you add to get from the 4th
4
20
term to the 20th term?
a  3, a  35
a
3520  a34  16d
Solve this for d
The fourth term is the first term
plus 3 common differences.
a1  3
d=2
3a4  a1  3d(2)
We have all the info we need to express
these sequences. We’ll do it on next slide.
The fourth term is 3 and the 20th term is 35. Find the first
term and both a term generating formula and a recursive
formula for this sequence.
a4  3, a20  35
d=2
a1  3
makes the first term - 3 instead of 2
2n  5
makes the common difference 2
a4  24  5  3
a20  220  5  35
Let’s check it out. If we find n = 4 we should get the 4th
term and n = 20 should generate the 20th term.
The recursive formula would be:
an  an1  2
Often in applications we will want the sum of a certain
number of terms in an arithmetic sequence.
The story is told of a grade school teacher In the 1700's
that wanted to keep her class busy while she graded
papers so she asked them to add up all of the numbers
from 1 to 100. These numbers are an arithmetic sequence
with common difference 1. Carl Friedrich Gauss was in
the class and had the answer in a minute or two
(remember no calculators in those days). This is what he
did:
sum is 101
1 + 2 + 3 + 4 + 5 + . . . + 96 + 97 + 98 + 99 + 100
sum is 101
With 100 numbers there are 50 pairs that add up to 101.
50(101) = 5050
This will always work with an arithmetic sequence. The
formula for the sum of n terms is:
n is the number of terms so n/2 would be the number of pairs

n
S n  a1  an
2
first term

last term
Let’s find the sum of 1 + 3 +5 + . . . + 59
But how many terms are there?
We can write a formula for the sequence and then figure
out what term number 59 is.

n
S n  a1  an
2
first term

last term
Let’s find the sum of 1 + 3 +5 + . . . + 59
The common difference is 2 and
the first term is one so:
2n 1
Set this equal to 59 to find n. Remember n is the term number.
2n - 1 = 59 n = 30
So there are 30 terms to sum up.
30
S30  1  59  900
2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au