Sequences - I - The Geometric Sequence
This section will provide an introduction to geometric sequences.
A geometic sequence is a sequence where we obtain each term after the first by multiplying the previous term by a common multiplier (frequently called the common ratio). We find the common ratio of a sequence by dividing any term (other than the first) by the preceeding term. For example we might have the sequence 3, 6, 12, 24, 48. .... If we divide 48 by 24 we get the common ratio = 48/24 = 2. Dividing 12/6 gives the same result. In many texts the common ratio is referred to by the letter r. The terms of the sequence are represented by a1, a2, a3 ... an. For every positive integer n we can define a term in the sequence by the relationship an+1 = r*al.
The general terms of a geometric sequence can we written as:
- First term, a1
- Second term, a2 = a1 * r
- Third term, a3 = a1 * r2
- nth term, an = a1 * rn-1
- The line above this one defines the general term of a geometric sequence
Individual Terms in a Sequence
To recap: When sequences are talked about in a general way, with symbols the terms of the sequence are describes as a1, a2... The ratio of sequential terms, the common ratio is usually represented by the letter r. So you can say that r = an+1/an. In other words r is equal to the ratio of adjacent terms in the sequence. Lets take a sequence and write some of the terms.
Geometric sequence with a1 = 5 and r = 2
- a1 = 5
- a2 = a1*r = 5*2 = 10
- a3 = a1*r3-1 = 5*22 = 5*4 = 20
- a4 = a1*r4-1 = 5*23 = 5*8 = 40
- .................................
- a10 = a1*r10-1 = 5*29 = 2560
- The equation we used to generate the terms was an = a1*rn-1
Let's use what was demonstrated above to solve a problem. Suppose you are given the geometric sequence 4, 12, 36, 108, ... and asked to find the expression for the general term of this sequence.
The general term is an = a1*rn-1 so we need to determine the values of a1 and r
- Since the first term in the given sequence is 4 we must conclude that a1 = 4
- Since this is a geometric sequence we can take the ratio of any two adjacent terms: 108/36 = 3 or 36/12 = 3 or 12/4 = 3
- If this is really a geometric sequence all of the ratios must give us the same value. In this problem that value is r = 3
- We now know a1 and r so we can write an = 4*3n-1
- NOTE: In general you CANNOT multiply the 4 times the 3. The 3 is taken to a power while the 4 is not and they are not the same base.
- To get an all numbers answer you need a value of n and the problem doesn't provide one, it just asks for this expression.
- If you were now asked to find the value of the 7th term in this sequence you could write:
- a7 = 4*37-1 = 4*36 = 4*729 = 2916
To help solidify this let's look at a few problems. We'll start simply and progress to the (impossibly) difficult.
Determining the type of sequence and the general expression when given a sequence
Sum of the Terms in a Geometric Sequence
Up to this point we have discussed the individual terms of a geometric sequence. The next question to address is the summing of all those individual terms to get the sum of the terms in a sequence. In the case of an arithmetic sequence we added a constant amount to a term to get the next term. For a geometric sequence we multiply a term by a constant to get the next term. Let's use that information to see what we can learn about sums of geometric sequences.
Lets next consider an geometric sequence such as 1, 2, 4, 8, 16, 32
We will use a sequence of 10 terms with a1 = 1 and r = 2
- If we let the sum = S we can write S = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512
- Now lets multiply S by r to get 2*S = 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024
- Next we calculate 2S - S = 1024 - 1 (all other terms cancel one another)
- Or S = 1023, the sum of our 10 term series is 1,023
- Now let's take a more general approach
- Sn = a1 + a1*r + a1*r2 + ......+ a1*rn-1
- Now we will multiply the above equation by r
- .r*Sn = a1*r + a1*r2 + a1*r3 + ......+ a1*rn
- Next we subtract the S equation from the r*S equation
- .r*Sn - Sn = a1*rn - a1 (all other terms cancel out)
- Factoring Sn gives us Sn*(r - 1) = a1*rn - a1
- Solving for Sn we get Sn = ( a1*rn - a1)/(r-1)
- This is our general formula for the sum of the terms of a geometric series of length n with common ratio r.
To start let's look at a problem that asks us to sum the first 7 terms of the geometric sequence ...2, 6, 18, 54, ...
..
a1 = 2 and r = 54/18 or 18/6 or 6/2 all of which give us r = 3
- Our formula says Sn = ( a1*rn - a1)/(r-1)
-
- Substituting, S10 = (2*310 - 2)/(3 - 1) = (2*59049 - 2)/2 = 59048
- The sum of the terms of our geometric series is 59,048
- Let's look at one more that is a little different
- Find the the sum of the terms in the series 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... 1/128
- We know a1 = 1 and we can determine that r = 1/2. But, we don't know the value of n
- We do know that the last term is 1/128. So let's take our formula for an and determine the value of n
- We know an = a1*rn-1>
- 1/128 = 1*(1/2)n-1
- 1/128 = (1/2)7. So it must be true that 7 = n-1, or n = 7 + 1 = 8
- Note: If you know logarithms you could also solve this problem by taking the logs of both sides of the equation
- Now S8 = (1*(1/2)8 - 1)/(1/2 - 1) = (1/128 - 128/128)/(-1/2)
- S8 = (-127/128)/-1/2) = 127/64 = 1.984375
- If r is less than 1 an alternative form of the formula that may be easier to use is:
- Sn = ( a1 - a1)*rn/(1- r) (we factored a -1 from numerator an denominator)
One question we can ask about geometric sequences is, "What happens to the sum of the terms as n approaches infinity?"
- Look at our formula for the sum of terms, Sn = ( a1 - a1)*rn/(1- r)
- We'll split the right-hand side of the equation into two terms Sn = a1(1-r) - a1)*rn/(1- r)
-
- If |r| < 1 then as n gets larger rn will get smaller. For example ir r = 1/2 we have 1/2, 1/4, 1/8, 1/16 etc....
- We can make rn as close to zero as we desire by carefully choosing values of n
- If |r| < 1 as n approaches infinity the term (a1*rn)/(1- r) will approach zero
- This leaves us with S ∞ = a1/(1- r)
- This is the formula for the sum of the infinite geometric sequence
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Determining the sum of the terms of a sequence
Word Problems
We'll end this page with a couple of word problems. To help you work these let me tell you a little bit about what is usually called exponential growth. Assume you invest $1000 in a mutual fund that pays 8% per year (long term average). Here is how the amount of money you have will change:
- a1 = $1,000
- a2 = 1000 + 1000*.08 = 1000(1+.08) = 1000*1.08
- a3 = 1000*1.08 + .08*[1000*1.08] = 1000*1.08*[1+.08] = 1000*1.082
- a4 = 1000*1.083
- an = 1000*1.08n-1
- Not by accident this is exactly like our formula for the nth term of a geometric sequence
- For an investment problem we can write Value of investment in year n = Initial Investment*(1 + interest rate as a decimal)time - 1;
Now let's look at a couple of word problems that involve investments.