Download Sec 13.1 Arithmethic and Geometric Sequences

13-1 Arithmetic and Geometric Sequences
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.
NOTATION
1st term = t₁
2nd
term = t₂
3rd term = t₃
4th term = t₄
nth term = tn
Some people use a to
describe the terms of a
sequence. However
some prefer to use t.
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
d=4
3, 7, 11, 15, 19 …
𝑎1 = 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.
t n  t  n  1d
Try this to get the 5th term.
t5  3  5  14  3  16  19
Let’s look at a formula for an arithmetic sequence and see what it tells us.
you can see what the common
difference will be in the
formula
4n 1
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.
Subbing in the set of positive integers we get:
3, 7, 11, 15, 19, …
What is the
common
difference?
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, . . .?
d=4
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.
Sure enough---it starts at 6 and has a
common difference of -2
6, 4, 2, 0, -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.
a4  3, a20  35
a
35
34  16d
20  a
The fourth term is the first term plus 3
common differences.
a1  3
How many differences would you add
to get from the 4th term to the 20th
term?
Solve this for d
d=2
a4  a1  3d(2)
3
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
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
𝑠𝑡𝑒𝑝 1: 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑛𝑐𝑒
35 − 3 32
𝑑=
=
=2
20 − 4 16
𝑆𝑡𝑒𝑝 2: 𝑢𝑠𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑡𝑒𝑟𝑚:
𝑎𝑛 = 2 𝑛 − 4 + 3
Simply: 𝑎𝑛 = 2𝑛 − 5
Or use the other term:
𝑎𝑛 = 2 𝑛 − 20 + 35
Simplify: 𝑎𝑛 = 2𝑛 − 5
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 +
sum is 101
100
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 + . . . +
But how59
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
GEOMETRIC
SEQUENCES
These are sequences where the ratio of
successive terms of a sequence is always
the same number. This number is called
the common ratio.
1, 2, 4, 8, 16 . . .
r=2
Notice in this sequence that if we find the ratio of any
term to the term before it (divide them) we always get 2.
2 is then called the common ratio and is denoted with the
letter r.
To get to the next term in the sequence we would
multiply by 2 so a recursive formula for this sequence is:
an  2an 1
𝑻𝒉𝒊𝒔 𝒊𝒔 𝒂
𝒓𝒆𝒄𝒖𝒓𝒔𝒊𝒗𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂.
2 2 2 
2
1, 2, 4, 8, 16 . . .
r=2
a=1
Each time you want another term in the sequence you’d multiply by r. This
would mean the second term was the first term times r. The third term is the
first term multiplied by r multiplied by r (r squared). The fourth term is the
first term multiplied by r multiplied by r multiplied by r (r cubed). So you can
see to get the nth term we’d take the first term and multiply r (n - 1) times.
an  ar
n 1
𝑻𝒉𝒊𝒔 𝒊𝒔 𝒂
𝒓𝒆𝒄𝒖𝒓𝒔𝒊𝒗𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂.
Try this to get the 5th term.
a5  12
51
 16
Let’s look at a formula for a geometric sequence and see what it tells us.
 23 
you can see what the common
ratio will be in the formula
n 1
This factor gets us started in
the right place. With n = 1 we’d
get -2 for the first term
Subbing in the set of positive integers we get:
-2, -6, -18, -54 …
What is the
common ratio?
3n-1 would generate the powers of 3. With the - 2 in front,
the first term would be
-2(30) =- 2. What would you do if you wanted the
sequence -4, -12, -36, -108, . . .?
r=3
 43 
n 1
Find the nth term of the geometric sequence when a = -2 and r =4
If we use 4n-1 we will generate a sequence whose common ratio is 4, but
this sequence starts at 1 (put 1 in for n to get first term to see this). We
want ours to start at -2. We then need the “compensating factor”. We
need to multiply by -2.
 24 
n 1
Check it out by putting in the first few positive integers and
verifying that it generates our sequence.
Sure enough---it starts at -2 and has a
common ratio of 4
-2, -8, -32, -128, . . .
Find the 8th term of 0.4, 0.04. 0.004, . . .
an  ar
To find the common ratio, take any
term and divide it by the term in
front
n 1
0.04
r
 0.1
0.4
an  0.40.1
n1
a8  0.40.1
81
 0.00000004
If we want to add n terms in a geometric sequence, we use the formula
below:
number of terms
first term
1 r
Sn  a
, r  0, 1
1 r
n
sum of n terms
15
Find the sum:
 43
n 1
common ratio
n 1
1 3
S15  4
1 3
15
4 + 12 + 36 + 108 + . . . + 4(3)14= 28,697,812
1
an   
2
Let’s look at the sum of the geometric sequence
1 1 1 1
   
2 4 8 16
1
2
0
n
Let’s look at this on the
number line

1
4

1 1

8 16
1
Each time we add another term we’d be going half the distance left. As n 
 the sum  1.
means infinity
If the common ratio was not a fraction between -1 and 1, then the sequence
would keep getting larger and larger and would   as n  . If the
common ratio is a fraction between -1 and 1, the sum as n   is as follows:

 ar
k 1
k 1
a

1 r
first term
common ratio
Let’s try this for the previous sequence:
1
1
S  2  2  1
1
1
1
2
2
1 1 1 1
   
2 4 8 16
1
x
2
1
x
2
1
x
2
Let’s try one more:
8a

1
8 

k 1  3 
8
  12
S 
1  r1 2
3
3
k 1
Example 1
Solution 1
Example 2
Solution 2
Example 3
In a geometric sequence:
𝑡3 = 12 𝑎𝑚𝑑 𝑡6 = 96
𝑎) 𝑓𝑖𝑛𝑑 𝑎 𝑡ℎ𝑒 𝑟𝑢𝑙𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒.
𝑏) 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 11𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑚𝑐𝑒.
Solution 3
Homework
Written exercises on Page 476: