ARITHMETIC SEQUENCES

1
ARITHMETICSEQUENCES
These are sequences where the difference between
successive terms of a sequence is always the same
number. This number is called the common
difference.
2
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
The first term in the sequence would be a1 which
is sometimes just written as a.
3
d 4
a 3
3, 7, 11, 15, 19
Each time you want another term in the sequence
youd 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 wed take the first term and add d (n -
1) times.
Try this to get the 5th term.
4
Lets look at a formula for an arithmetic
sequence and see what it tells us.
We can think of this as a compensating term.
Without it the sequence would start at 4 but this
gets it started where we want it.
you can see what the common difference will be in
the formula
Subbing in the set of positive integers we get
What is the common difference?
3, 7, 11, 15, 19,
d 4
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, . . .?
5
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, wed need to add 8.
Check it out by putting in the first few positive
integers and verifying that it generates our
sequence.
Sure enough---it starts at 6 and has a common
difference of -2
6, 4, 2, 0, -2, . . .
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Lets try something a little trickier. What if
we just know a couple of terms and they arent
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 term to the 20th term?
d 2
35
3
Solve this for d
(2)
The fourth term is the first term plus 3 common
differences.
3
We have all the info we need to express these
sequences. Well do it on next slide.
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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.
d 2
makes the first term - 3 instead of 2
makes the common difference 2
Lets 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
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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
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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
last term
first term
Lets 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.
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last term
first term
Lets find the sum of 1 3 5 . . . 59
The common difference is 2 and the first term is
one so
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.
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Acknowledgement I wish to thank Shawna Haider
from Salt Lake Community College, Utah USA for
her hard work in creating this PowerPoint. www.sl
cc.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
Stephens School Carramar www.ststephens.wa.edu.
au