Sequences - I - The Arithmetic Sequence
Many college math texts assume that you clearly remember the nature of sequences and how the formulas for handling them were developed when you took a basic algebra course. My experience tells me that most of you only have a vague recollection of sequences and have no idea how the formulas are derrived. So, this section will start with the basics on arithmetic sequences and go through the process of working with them. The following section will deal with geometric sequences. If you have both of these down pat you will not get much out of this section of the website.
If you are still reading you probably don't remember much about arithmetic sequences or may not have even come across them before.
An arithmetic sequence is a sequence where you have a common (constant) difference between successive terms.
For example
- 3, 5, 7, 9,..... is an arithmetic sequence where the difference between successive terms is 2
- 6, 2, -4, -8... is an arithmetic sequence where the difference between successive terms is -4 (the difference doesn't have to be positive)
- 2, 5, 9, 12, 15... is NOT an arithmetic sequence because the difference between terms is not constant. Sometimes it is 4 and sometimes it is 3
Individual Terms in a Sequence
When sequences are talked about in a general way, with symbols the terms of the sequence are describes as a1, a2...
The difference between terms, the common difference is usually represented by the letter d. So you would say that d = an+1 - an. In other words d is equal to the difference between two adjacent terms in the sequence. If we are working with a sequence we can write the terms as follows:
Terms in arithmetic sequence
- First term a1
- Second term a2 = a1 + d (we have added the commom difference)
- Third term a3 = a1+d + d = a1 + 2d
- Fourth term a4 = a1+d + d + d = a1 + 3d
- nth term an = a1 + (n-1)d
- The equation shown can be used to represent any term in an arithmetic sequence, it is sometimes called the general term.
Let's use what was discussed above. We are given the sequence 7, 3, -1, -5, ... The question then asks us to find the common difference.
- if you use the last two terms you have -5 - (-1) = -4
- if you use the two middle terms you have -1 -(+3) = -4
and finally if you used the first two terms you have 3 - 7 = -4
- Note it is always an+1 - an the later term minus the earlier term
- Now suppose we are asked to write the general term for this sequence
- We know a1 = 7 and we know d = 7 and we know the nth term an = a1 + (n-1)d
- So, we can write an = 7 + (n - 1)*(-4) This is the answer we have been asked to find.
- We could simplify it further by distributing the -4 and combining like terms = 7 -4n + 4 = -4n + 11
- Whether you would do this depends on the form of the answer that your instructor wants
- Instructors being the way they are suppose your instructor now asks you to find the value of the 11th term in the sequence.
- You can write a11 = 7 + (11 - 1)*(-4) = 7 + (10)*(-4) = 7 - 40 = -33.
- The value of the 11th term is -33.
To help solidify this let's look at a few problems. We'll start simply and progress to the (impossibly) difficult.
Determining the general expression when given a sequence
Sum of the Terms in a Sequence
Up to this point we have discussed the individual terms of a sequence. The next question to address is the summing of all those individual terms to get the sum of the terms in a sequence. Let me first introduce the Greek capital letter sigma written as Σ. When you see this letter it should tell you to sum all of the terms that follow it. Usually it is written with a lower limit, on the bottom and an upper limit, on the top. Here is an example (yes, its tiny and hard to read). This tells us that i goes from 1 to 5 inclusive and asks us to add the terms 1, 2, 3, 4 and 5. In other words it says total 1 + 2 + 3 + 4 + 5. The answer being 15. We have found the sum of an arithmetic sequence of 5 terms. The first term a1 was equal to 1 and the difference, d, was equal to 1.
Lets next consider an arithmetic sequence that goes from 1 to 10 with a difference between terms of 1.
- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- Now lets write the sequence backwords, from 10 down to 1 (a perfectly legitimate thing to do)
- 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
- now let's add terms vertically
- 11, 11, 11, 11, 11, 11, 11, 11, 11, 11
- We got ten 11's. Both rows represent the same series, so 10*11 represents twice the sum of the sequence or 2*Sn
- Sn is typically used to represent the sum of a sequence.
- What we know is 2*Sn
= 10*11 so Sn = (10*11)/2 = 5*11 = 55
- 55 represents the sum of all of the terms in the sequence and we did it without adding all of the terms individually.
- -------------Can we use this information to write a general formula ? (of course we can)-----------------
- We could have written the sequence as a1 + (a1 + d) + (a1 + 2d) + .... + (an - d) + an
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- Backwards we would have written an + (an - d) + ... (a1 + d) + a1
- Then, when we added the two we would have 2*Sn = (a1 + an) + (a1 + an) + ....
- Or 2*Sn = n*(a1 + an)
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- Which leads us to the formula Sn = n*(a1 + an)/2
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- This is the formula for the sum of an aruthmatic sequence
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- We will now use the formula to solve some problems
To start let's look at a problem that asks us to sum the first thirty terms of the arithmetic sequence 3, 7, 11, 15 ..
..
We want Sn = but we don't want to manually add 30 terms
- Our formula says Sn = n*(a1 + an/2 so we need the values of n, a1 and an
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- We know that a1 = 3 because that is given to us in the statement of the problem
- We know that d = 7-3 or 11-7 or 15-11 all of which equal 4.
- The formula for the nth term (from above) is an = a1 + (n-1)*d
- So a30 = 3 + (30-1)*4 = 3 + 29*4 = 3 + 116 = 119
- Now we can find Sn since Sn= n*(a1 + an)/2 = 30*(3 + 119)/2 = 1830.
- The sum of the first 30 terms of our sequence is 1830.
Let's look at a slightly different version of the problem. Suppose we are given the sequence 7 + 10 + 13 + ... + 157 and asked to find the sum of the sequence.
We know the first term is 7 the last term is 157 and the difference between terms in 3 but we don't know n the number of terms in the sequence. How can we find n? Well we have a formula that says an = a1 + (n-1)*d and we know everything in this formula except n. So we can use it to find n.
- 157 = 7 + (n-1)*3
- 157 = 7 + 3n - 3
- 157 = 3n + 4
- 3n = 157 - 4 = 153
- and finally n = 153/3 = 51.
- Our sequence has 51 terms.
- Now using our sum formula we have Sn= 51*(7 + 157)/2 = 4182.
Determining the sum of the terms of a sequence
Word Problems
Let's end this page with a couple of word problems.