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Introduction 


Sequences of numbers that follow a pattern of adding a fixed number from one term to the next are called arithmetic
sequences. The following sequences are arithmetic sequences:
Sequence A: 5 , 8 , 11 , 14 , 17 , ... Sequence B: 26 , 31 , 36 , 41 , 46 , ... Sequence
C: 20 , 18 , 16 , 14 , 12 , ...
For sequence A, if we add 3 to the first number we will get the second number. This works for any pair of consecutive numbers.
The second number plus 3 is the third number: 8 + 3 = 11, and so on.
For sequence B, if we add 5 to the first number we will get the second number. This also works for any pair of consecutive
numbers. The third number plus 5 is the fourth number: 36 + 5 = 41, which will work throughout the entire sequence.
Sequence C is a little different because we need to add 2 to the first number to get the second number. This too works
for any pair of consecutive numbers. The fourth number plus 2 is the fifth number: 14 + (2) = 12.
Because these sequences behave according to this simple rule of addiing a constant number to one term to get to another,
they are called arithmetic sequences. So that we can examine these sequences to greater depth, we must know that the fixed
numbers that bind each sequence together are called the common differences. Sometimes mathematicians use the letter d
when referring to these types of sequences.
Mathematicians also refer to generic sequences using the letter a along with subscripts that correspond to the term
numbers as follows:
Generic Sequence: a_{1}, a_{2}, a_{3}, a_{4}, ...
This means that if we refer to the fifth term of a certain sequence, we will label it a_{5}. a_{17} is
the 17th term. This notation is necessary for calculating nth terms, or a_{n}, of sequences.
d can be calculated by subtracting any two consecutive terms in an arithmetic sequence.
d = a_{n}  a_{n  1}, where n is any positive integer greater than 1.


In order for us to know how to obtain terms that are far down these lists of numbers, we need to develop a formula that
can be used to calculate these terms. If we were to try and find the 20th term, or worse to 2000th term, it would take a long
time if we were to simply add a number  one at a time  to find our terms.
If a 5yearold was asked what the 301st number is in the set of counting numbers, we would have to wait for the answer
while the 5yearold counted it out using unnecessary detail. We already know the number is 301 because the set is extremely
simple; so, predicting terms is easy. Upon examining arithmetic sequences in greater detail, we will find a formula for each
sequence to find terms.
 Let's examine sequence A so that we can find a formula to express its nth term.
If we match each term with it's corresponding term number, we get:
n 
1 
2 
3 
4 
5 
. . . 
Term 
5 
8 
11 
14 
17 
. . . 
The fixed number, called the common difference (d), is 3; so, the formula will be a_{n} = dn + c or a_{n}
= 3n + c, where c is some number that must be found.
For sequence A above, the rule a_{n} = 3n + c would give the values... 3×1 + c = 3 + c 3×2 + c = 6 + c 3×3
+ c = 9 + c 3×4 + c = 12 + c 3×5 + c = 15 + c
If we compare these values with the ones in the actual sequence, it should be clear that the value of c is 2. Therefore
the formula for the nth term is...
a_{n} = 3n + 2.
Now if we were asked to find the 37th term in this sequence, we would calculate for a_{37} or 3(37) + 2 which is
equal to 77 + 2 = 79. So, a_{37} = 79, or the 79th term is 79. Likewise, the 435th term would be a_{435} =
3(435) + 2 = 1305.
 Let's take a look at sequence B.
n 
1 
2 
3 
4 
5 
. . . 
Term 
26 
31 
36 
41 
46 
. . . 
The fixed number, d, is 5. So the formula will be a_{n} = dn + c or a_{n} = 5n + c .
For the sequence above, the rule a_{n} = 5n + c would give the values...
5×1 + c = 5 + c 5×2 + c = 10 + c 5×3 + c = 15 + c 5×4 + c = 20 + c 5×5 + c = 25 + c
If we compare these values with the numbers in the actual sequence, it should be clear that the value of c is 21. Therefore,
the formula for the nth term is...
a_{n} = 5n + 21.
If we wanted to calculate the 14th term, we would calculate for
a_{14} = 5(14) + 21 = 70 + 21 = 91. If we needed the 40th term, we would calculate a_{40} = 5(40) + 21
= 200 + 21 = 221. The general formula is very handy.
 Now let's do the third and final example....
n 
1 
2 
3 
4 
5 
. . . 
Term 
20 
18 
16 
14 
12 
. . . 
The common difference is 2. So the formula will be 2n + c, where c is a number that must be found.
For sequence C, the rule 2n + c would give the values...
2×1 + c = 2 + c 2×2 + c = 4 + c 2×3 + c = 6 + c 2×4 + c = 8 + c 2×5 + c = 10 + c
If we compare these values with the numbers in the actual sequence, it should be clear that the value of c is 22. Therefore,
the formula for the nth term is...
a_{n} = 2n + 22.
If for some reason we needed the 42nd term, we would calculate for a_{42} = 2(42) + 22 = 84 + 22 = 62. Similarly,
a_{90} = 2(90) + 22 = 180 + 22 = 158.


It may be necessary to calculate the number of terms in a certain arithmetic sequence. To do so, we would
need to know two things.
We would need to know a few terms so that we could calculate the common difference and ultimately the formula for the general
term. We would also need to know the last number in the sequence.
Once we know the formula for the general term in a sequence and the last term, the procedure is relatively uncomplicated.
Set them equal to each other. Since the formula uses the variable n to calculate terms, we can also use it to determine
the term number for any given term.
 If we again look at sequence A above, let's use the formula that was found to calculate term values, a_{n} = 3n
+ 2. If we knew that 47 was a number in the sequence  5, 8, 11, 14, 17, ..., 47  we would set the number 47 equal to the
formula a_{n} = 3n + 2, we would get 47 = 3n + 2. Solving this equation yields n = 15. This means that there are 15
terms in the sequence and that the 15th term, a_{15}, is equal to 47.
 Let's look at a portion of sequence C. If the sequence went from 20 to 26, we would have: 20, 18, 16, 14, 12, ...,26.
We would use the formula for the general term, a_{n} = 2n + 22, and set it equal to the last term, 26. We would
get 26 = 2n + 22 and algebra would allow us to arrive at n = 24. This means that there are 24 terms in the sequence and
that a_{24} = 26.


Given our generic arithmetic sequence a_{1}, a_{2}, a_{3}, a_{4}, ..., we
can add the terms, called a series, as follows: a_{1} + a_{2} + a_{3} + a_{4} + ... + a_{n}.
Given the formula for the general term a_{n} = dn + c, there exists a formula that can add such a finite list of these
numbers. It requires three pieces of information. The formula is...
S_{n} = ½n(a_{1} + a_{n})
...where S_{n} is the sum of the first n numbers, a_{1} is the first number in the sequence
and a_{n} is the last number in the sequence.
Usually problems present themselves in either of two ways. Either the first number in the sequence and the number
of terms are known or the first number and the last number of the sequence are known.
 Let's take a finite portion of sequence B and experience our first case. If we had 26, 31, 36, 41, 46, ... and knew that
there were 50 terms in the sequence, then we have a_{1} = 26 and n = 50. We would have to develop a formula for the
nth term so we could calculate a_{50}, the last term in the sequence. Since we already calculated the formula above,
we can use it to calculate a_{50}. a_{n} = 2n + 22 is the formula so a_{50} = 2(50) + 22 = 100
+ 22 = 78. Now we can plug the numbers into the formula and gain a solution. S_{50} = ½(50)(26 + 78) = 25(52) =
1300. This means that the sum of the first 50 terms is 1300.
 Next, let's take a portion of sequence A for our second possible situation. If we were dealing with 5, 8, 11, 14, 17,
... , 231, then we would know that a_{1} = 5 and a_{n} = 231. If we knew the number of terms in this sequence,
we would be able to use the formula. Finding n becomes our next task. Since we know the formula for the general term,
a_{n} = 3n + 2, we can use it to find the number of terms in this sequence. We set the last term equal to the formula
and solve for n. We get 231 = 3n + 2, which means that n = 42 and a_{42} = 231. Now we can plug the information into
the sum formula and get S_{42} = ½(42)(5 + 231) = (21)(236) = 4956, which must be the sum of the first 42 terms in
the sequence.


Introduction 


Sequences of numbers that follow a pattern of multiplying a fixed number from one term to the next are called
geometric sequences. The following sequences are geometric sequences:
Sequence A: 1 , 2 , 4 , 8 , 16 , ... Sequence B: 0.01 , 0.06 , 0.36 , 2.16 , 12.96 , ... Sequence
C: 16 , 8 , 4 , 2 , 1 , ...
For sequence A, if we multiply by 2 to the first number we will get the second number. This works for any pair of consecutive
numbers. The second number times 2 is the third number: 2 × 2 = 4, and so on.
For sequence B, if we multiply by 6 to the first number we will get the second number. This also works for any pair of
consecutive numbers. The third number times 6 is the fourth number: 0.36 × 6 = 2.16, which will work throughout the entire
sequence.
Sequence C is a little different because it seems that we are dividing; yet to stay consistent with the theme of geometric
sequences, we must think in terms of multiplication. We need to multiply by 1/2 to the first number to get the second number.
This too works for any pair of consecutive numbers. The fourth number times 1/2 is the fifth number: 2 × 1/2 = 1.
Because these sequences behave according to this simple rule of multiplying a constant number to one term to get to another,
they are called geometric sequences. So that we can examine these sequences to greater depth, we must know that the fixed
numbers that bind each sequence together are called the common ratios. Mathematicians use the letter r when referring
to these types of sequences.
Mathematicians also refer to generic sequences using the letter a along with subscripts that correspond to the term
numbers as follows:
Generic Sequence: a_{1}, a_{2}, a_{3}, a_{4}, ...
This means that if we refer to the tenth term of a certain sequence, we will label it a_{10}. a_{14} is
the 14th term. This notation is necessary for calculating nth terms, or a_{n}, of sequences.
r can be calculated by dividing any two consecutive terms in a geometric sequence.
r = a_{n}/a_{n  1}, where n is any positive integer greater than 1


In order for us to know how to obtain terms that are far down these lists of numbers, we need to develop a formula that
can be used to calculate these terms. If we were to try and find the 20th term, or worse to 2000th term, it would take a long
time if we were to simply multiply a number  one at a time  to find our terms.
If we had to find the 400th term of sequence A above, we would undertake a tedious task had we decided to multiply by two
each step of the way all the way to the 400th term. Luckily, there is a way to arrive at the 400th term without the need for
calculating terms 1 through 399.
The formula for the general term for each geometric sequence is...
a_{n} = a_{1}r^{n  1}
 Let's examine sequence A so that we can find a formula to express its nth term.
If we match each term with it's corresponding term number, we get:
n 
1 
2 
3 
4 
5 
. . . 
Term 
1 
2 
4 
8 
16 
. . . 
The fixed number, called the common ratio (r), is 2; so, the formula will be a_{n} = a_{1}2^{n  1}
or a_{n} = (1)2^{n  1} or...
a_{n} = 2^{n  1}
For sequence A above, the rule a_{n} = 2^{n  1} would give the values... 2^{1  1} = 2^{0}
= 1 2^{2  1} = 2^{1} = 2 2^{3  1} = 2^{2} = 4 2^{4  1} = 2^{3}
= 8 2^{5  1} = 2^{4} = 16
Now if we were asked to find the 12th term in this sequence, we would calculate for a_{12} or 2^{12  1}
which is equal to 2^{11} = 2048. So, a_{11} = 2048, or the 11th term is 2048. Likewise, the 20th term would
be a_{20} = 2^{19} = 524288.
 Let's take a look at sequence B.
n 
1 
2 
3 
4 
5 
. . . 
Term 
0.01 
0.06 
0.36 
2.16 
12.96 
. . . 
The fixed number, r, is 6. So the formula will be...
a_{n} = (0.01)6^{n  1} For the sequence above, the rule a_{n} = (0.01)6^{n
 1} would give the values...
(0.01)6^{1  1} = (0.01)6^{0} = (0.01)(1) = 0.01 (0.01)6^{2  1} = (0.01)6^{1} = (0.01)(6)
= 0.06 (0.01)6^{3  1} = (0.01)6^{2} = (0.01)(36) = 0.36 (0.01)6^{4  1} = (0.01)6^{3}
= (0.01)(216) = 2.16 (0.01)6^{5  1} = (0.01)6^{4} = (0.01)(1296) = 12.96
If we wanted to calculate the 8th term, we would calculate for
a_{8} = (0.01)6^{8  1} = (0.01)6^{7} = (0.01)(279936) = 2799.36. If we needed the 12th term, we
would calculate a_{12} = (0.01)6^{12  1} = (0.01)6^{11} = (0.01)(362797056) = 3627970.56. As one
can see, the general formula is very handy when n is large.
 Now let's do the third and final example with sequence C....
n 
1 
2 
3 
4 
5 
. . . 
Term 
16 
8 
4 
2 
1 
. . . 
The common ratio is 1/2. So the formula will be ...
a_{n} = (16)(1/2)^{n  1} For sequence C, the rule a_{n} = (16)(1/2)^{n
 1} would give the values...
(16)(1/2)^{1  1} = (16)(1/2)^{0} = (16)(1) = 16 (16)(1/2)^{2  1} = (16)(1/2)^{1}
= (16)(1/2) = 8 (16)(1/2)^{3  1} = (16)(1/2)^{2} = (16)(1/4) = 4 (16)(1/2)^{4  1} = (16)(1/2)^{3}
= (16)(1/8) = 2 (16)(1/2)^{5  1} = (16)(1/2)^{4} = (16)(1/16) = 1
If for some reason we needed the 11th term, we would calculate for a_{11} = (16)(1/2)^{11  1} = (16)(1/2)^{10}
(16)(0.0009765625) = 0.015625. Similarly, a_{16} = (16)(1/2)^{16  1} = (16)(1/2)^{15} = (16)(0.000030517578125)
= 0.00048828125.


It may be necessary to calculate the number of terms in a certain geometric sequence. To do so, we would need
to know two things.
We would need to know a few terms so that we could calculate the common ratio and ultimately the formula for the general
term. We would also need to know the last number in the sequence.
Once we know the formula for the general term in a sequence and the last term, the procedure is relatively uncomplicated,
but the technique needed to complete the procedure is complicated.
Set the last term equal to the formula for the general term. Since the formula uses the variable n to calculate
terms, we can also use it to determine the term number for any given term.
If we again look at sequence A above, let's use the formula that was found to calculate term values, a_{n} = a_{1}r^{n
 1}. If we knew that 256 was a number in the sequence (1, 2, 4, 8, 16, ..., 256 ) we would set the number 256 equal
to the formula a_{n} = a_{1}r^{n  1} and get 256 = 2^{n  1}. Solving this equation using
proper techniques requires the use of logarithms and would yield n = 9. [We could use the method of guessing and checking
to arrive at the same value.] This means that there are 9 terms in the mentioned sequence and that the 9th term, a_{9},
is equal to 256.
Let's look at a portion of sequence C. If the sequence went from 16 to 1/8, we would have: 16, 8, 4, ..., 1/8. We would
use the formula for the general term, a_{n} = (16)(1/2)^{n  1}, and set it equal to the last term, 1/8.
We would get 1/8 = (16)(1/2)^{n  1}. Solving this equation [with the use of either logarithms or the method of
guessing and checking] allows us to arrive at n = 8. This means that there are 8 terms in the sequence and that a_{8}
= 1/8.


Given our generic arithmetic sequence a _{1}, a _{2}, a _{3}, a _{4}, ..., a _{n},
we can look at it as a series: a _{1} + a _{2} + a _{3} + a _{4} + ... + a _{n}. There
exists a formula that can add a finite list of these numbers and a formula for an infinite list of these numbers. They both
require two pieces of information. Here are the formulas...
S_{n} = [a_{1}(1  r^{n})]/(1  r) for a finite geometric sequence, or
S_{infinite}
= a_{1}/(1  r) for infinite sequences only when 1 < r < 1
...where S_{n} is the sum of the first n numbers, a_{1} is the first number in the sequence and r is the
common ratio of the sequence. Let's use examples to investigate both formulas.
Formula One: Finite Sum
 If we wanted to add the first 7 terms of sequence A, we would know that a_{1} = 1, r = 2 and n = 7. We would then
plug those numbers into the formula and get S_{n} = [a_{1}(1  r^{n})]/(1  r) = [(1)(1  2^{7})]/(1
 2) = (1  128)/(1) = (127)/(1) = 127. So, S_{7} = 127.
 If we wanted to add the first 10 terms of sequence B, we would know that a_{1} = 0.01, r = 6 and n = 10. We would
then plug those numbers into the formula and get S_{n} = [a_{1}(1  r^{n})]/(1  r) = [(0.01)(1 
6^{10})]/(1  60466176) = (0.01)(60466175)/(5) = (604661.75)/(5) = 120932.35. So, S_{10} = 120932.35.
Formula Two: Infinite Sum
 If we had to add an infinite number of terms of sequence A, we would first see if it met the requirement of the formula
for the common ratio, r. If r is between 1 and 1, an infinite term sum exists. Since sequence A has an rvalue of
2 and 2 falls outside the acceptable range, there is no finite sum to that infinite list of numbers.
 Let's take an infinite number of terms from sequence C. If we had 16 + (8) + 4 + (2) + 1 + ..., then we would know that
a_{1} = 16 and r = 1/2. Since r lies within the acceptable limits, a finite sum exists. Using the infinite term sum
formula, we get S_{infinite} = a_{1}/(1  r) = 16/(1  (1/2)) = 16/1.5 = 10 2/3.
 If we had to add the infinite sum 27 + 18 + 12 + 8 + ..., we would first identify the first term a_{1}
and the common ratio r, which are 27 and 2/3 respectively. Since r lies between 1 and 1, the formula will work
and a finite sum exists for the infinite series. The formula yields S_{infinite} = 27/(1  (2/3)) = 27/(1/3) = 81.
 










