Rachel and Elaina have started a website where they debate the best color of hair dye. The show is really popular and the visitors to their website are increasing very rapidly. They figure that memberships are increasing by about 500 people every three days.
At this rate, how many members will they have on the 48th day? How many days will it be before they reach 25,000 members?
Watch This
Arithmetic Sequences: Watch this video on arithmetic sequences at this link.  Geometric Sequences: Learn about geometric sequences by watching the video at this link. 

Guidance
When we represent a sequence with a formula that lets us find any term in the sequence without knowing any other terms, we are representing the sequence explicitly.
Given a recursive definition of an arithmetic or geometric sequence, you can always find an explicit formula, or an equation to represent the n ^{ th } term of the sequence. Consider for example the sequence of odd numbers we started with: 1,3,5,7,...
We can find an explicit formula for the n ^{ th } term of the sequence if we analyze a few terms:
 a _{ 1 } = 1
 a _{ 2 } = a _{ 1 } + 2 = 1 + 2 = 3
 a _{ 3 } = a _{ 2 } + 2 = 1 + 2 + 2 = 5
 a _{ 4 } = a _{ 3 } + 2 = 1 + 2 + 2 + 2 = 7
 a _{ 5 } = a _{ 4 } + 2 = 1 + 2 + 2 + 2 + 2 = 9
 a _{ 6 } = a _{ 5 } + 2 = 1 + 2 + 2 + 2 + 2 + 2 = 11
Note that every term is made up of a 1, and a set of 2’s. How many 2’s are in each term?

a _{ 1 } = 1 a _{ 2 } = 1 + 2 = 3 a _{ 3 } = 1 + 2 × 2 = 5 a _{ 4 } = 1 + 3 × 2 = 7 a _{ 5 } = 1 + 4 × 2 = 9 a _{ 6 } = 1 + 5 × 2 = 11
The n ^{ th } term has ( n  1) 2's. For example, a _{ 99 } = 1 + 98 × 2 = 197 . We can therefore represent the sequence as a _{ n } = 1 + 2( n  1). We can simplify this expression:

a _{ n } = 1 + 2( n  1) a _{ n } = 1 + 2 n  2 a _{ n } = 2 n  1
In general, we can represent an arithmetic sequence in this way, as long as we know the first term and the common difference, d . Notice that in the previous example, the first term was 1, and the common difference, d , was 2. The n ^{ th } term is therefore the first term, plus d ( n  1):

a _{ n } = a _{ 1 } + d ( n  1)
You can use this general equation to find an explicit formula for any term in an arithmetic sequence.
Example A
Find an explicit formula for the nth term of the sequence 3, 7, 11, 15... and use the equation to find the 50 ^{ th } term in the sequence.
Solution
a _{ n } = 4 n  1 , and a _{ 50 } = 199
The first term of the sequence is 3, and the common difference is 4.

a _{ n } = a _{ 1 } + d ( n  1) a _{ n } = 3 + 4( n  1) a _{ n } = 3 + 4 n  4 a _{ n } = 4 n  1 a _{ 50 } = 4(50)  1 = 200  1 = 199
We can also find an explicit formula for a geometric sequence. Consider again the sequence in example 2a:

t _{ 2 } = 2 t _{ 1 } = 2 × 3 = 6 t _{ 1 } = 3 \begin{align*}\rightarrow\end{align*} t _{ 3 } = 2 t _{ 2 } = 2 × 6 = 12 t _{ n } = 2 × t _{ n } _{ 1 } t _{ 4 } = 2 t _{ 3 } = 2 × 12 = 24 t _{ 5 } = 2 t _{ 4 } = 2 × 24 = 48
Notice that every term is the first term, multiplied by a power of 2. This is because 2 is the common ratio for the sequence.

t _{ 1 } = 3 t _{ 2 } = 2 × 3 = 6 t _{ 3 } = 2 × 2 × 6 = 2 ^{ 2 } × 6 = 12 t _{ 4 } = 2 × 2 × 2 × 6 = 2 ^{ 3 } × 6 = 24 t _{ 5 } = 2 × 2 × 2 × 2 × 6 = 2 ^{ 4 } × 6 = 48
The power of 2 in the n ^{ th } term is ( n 1). Therefore the n ^{ th } term in this sequence can be defined as: t _{ n } = 3(2 ^{ n } ^{  1 } ). In general, we can define the n ^{ th } term of a geometric sequence in terms of its first term and its common ratio, r :
t _{ n }  = t _{ 1 } (r ^{ n } ^{ 1 } ) 

You can use this general equation to find an explicit formula for any term in a geometric sequence.
Example B
Find an explicit formula for the n ^{ th } term of the sequence 5, 15, 45, 135... and use the equation to find the 10 ^{ th } term in the sequence.
Solution
a _{ n } = 5 × 3 ^{ n } ^{  1 } , and a _{ 10 } = 98,415
 The first term in the sequence is 5, and r = 3.

a _{ n } = a ^{ 1 } × r ^{ n } ^{  1 } a _{ n } = 5 × 3 ^{ n  1 } a _{ 10 } = 5 × 3 ^{ 10  1 } a _{ 10 } = 5 × 3 ^{ 9 } = 5 × 19,683 = 98,415
Again, it is always possible to write an explicit formula for terms of an arithmetic or geometric sequence. However, you can also write an explicit formula for other sequences, as long as you can identify a pattern. To do this, you must remember that a sequence is a function, which means there is a relationship between the input and the output. That is, you must identify a pattern between the term and its index , or the term’s “place” in the sequence.
Example C
Write an explicit formula for the nth term of the sequence 1, (1/2), (1/3), (1/4)...
Solution
a _{ n } = (1/ n )
Initially you may see a pattern in the fractions, but you may also wonder about the first term. If you write 1 as (1/1), then it should become clear that the n ^{ th } term is (1/ n ).
Concept question wrapup "They figure that memberships are increasing by about 500 people every three days. At this rate, how many members will they have on the 48th day? How many days will it be before they reach 25,000 members?" This is actually a fairly simple arithmetic sequence: each day there are 500/3 more members, on average. Use the formula for arithmetic sequences from the first example above. 

Vocabulary
An explicit formula for a sequence allows you to find the value of any term in the sequence.
The natural numbers are a subset of the integers: {1,2,3,4,5....}
A recursive formula for a sequence allows you to find the value of the n ^{ th } term in the sequence if you know the value of the (n1) ^{ th } term in the sequence.
A sequence is an ordered list of objects.
Guided Practice
Questions
1) Write an explicit formula for the sequence: 2, 9, 16... and use the formula to find the value of the 20 ^{ th } term.
2) Write an explicit formula for the sequence: 5, 10, 20... and use the formula to find the value of the 9 _{ th } term.
3) Write an explicit formula for the sequence: (1/2), (1/4), (1/8) and use the formula to find the value of the 7 _{ th } term.
4) Identify all sequences in the previous three problems that are geometric. What is the common ratio in each sequence?
5) The membership of an online dating service increases at an average rate of 8% per year. In the first year, there are 500 members.
 a. How many members are there in the second year?
 b. How many members are there in the eighteenth year?
Solutions
1) For the sequence: 2, 9, 16...
 @$\begin{align*}a_n = 7n  5\end{align*}@$
 @$\begin{align*}\therefore a_{20} = 7(20)  5\end{align*}@$
 @$\begin{align*}a_{20} = 135\end{align*}@$
2) For the sequence: 5, 10, 20...
 @$\begin{align*}a_n = 5 \cdot 2^{n1}\end{align*}@$
 @$\begin{align*}\therefore a_9 = 5 \cdot 2^8\end{align*}@$
 @$\begin{align*}a_9 = 5 \cdot 256\end{align*}@$
 @$\begin{align*}a_9 = 1280\end{align*}@$
3) For the sequence: (1/2), (1/4), (1/8)...
 @$\begin{align*}a_n = \frac{1}{2^n}\end{align*}@$
 @$\begin{align*}\therefore a_7 = \frac{1}{2^7}\end{align*}@$
 @$\begin{align*}a_7 = \frac{1}{128}\end{align*}@$
4) The sequence in question 1 is arithmetic.
The sequence is question 2 is geometric, and has r = 2.
The sequence in question 3 is geometric and has r = 1/2.
5) a. 540 members
 b. Approximately 1,998 members
Practice
Name the sequence as arithmetic, geometric, or neither.
 @$\begin{align*}21, 6, 18, 3, 20, 2\end{align*}@$
 @$\begin{align*}0,\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, 1\end{align*}@$
 @$\begin{align*}1, 3, 9, 27, 81, 243\end{align*}@$
 @$\begin{align*}2, 9, 2, 1, 18, 2\end{align*}@$
Write the first 5 terms of the arithmetic sequence(explicit).
 @$\begin{align*}a_n = 8 9(n1)\end{align*}@$
 @$\begin{align*}a_n = 6\frac{2}{3}(n1)\end{align*}@$
 @$\begin{align*}a_n = 8 + \frac{1}{3}(n1)\end{align*}@$
Solve the following:
 What are the first five terms of the sequence? @$\begin{align*}a_n = a_{n1}  \frac{10}{3}; a_1 = 6\end{align*}@$
 Given the sequence, write a recursive function to generate it: @$\begin{align*}2, 4, 10, 16, 22, 28\end{align*}@$
 Write the equation of @$\begin{align*}a_n\end{align*}@$ without using recursion: @$\begin{align*}a_n = a_{n1}  \frac{3}{2}; a_1 = 10\end{align*}@$
 Write as a recursion: @$\begin{align*}a_n = 6  \frac{5}{3}(n1)\end{align*}@$
 Write the equation of @$\begin{align*}a_n\end{align*}@$ without using recursion: @$\begin{align*}a_n = a_{n1}+8; a_1 = 3\end{align*}@$
 What are the first five terms of the sequence? @$\begin{align*}a_n = a_{n1}  1; a_1 = 5\end{align*}@$
Write the formula for the explicit @$\begin{align*}(n_{th})\end{align*}@$ term of the arithmetic sequence.
 @$\begin{align*}7, \frac{13}{3}, \frac{5}{3}, 1, \frac{11}{3}, \frac{19}{3}\end{align*}@$
 @$\begin{align*}6, 4, 14, 24, 34, 44\end{align*}@$
 @$\begin{align*}9, 16, 23, 30, 37, 44\end{align*}@$
 In a particular arithmetic sequence, the second term is 4 and the fifth term is 13. Write an explicit formula for this sequence.
Write the first 5 terms of the geometric sequence
 @$\begin{align*}a_n = 5(3)^{(n1)}\end{align*}@$
 @$\begin{align*}a_n = 6(\frac{10}{3}^{(n1)})\end{align*}@$
Write the formula for the @$\begin{align*}n_{th}\end{align*}@$ term of the geometric sequence
 @$\begin{align*}8, 16, 32, 64, 128, 256\end{align*}@$
 @$\begin{align*}9, \frac{27}{2}, \frac{81}{4}, \frac{243}{8}, \frac{729}{16}, \frac{2187}{32}\end{align*}@$
Convert the explicit and rewrite as recursion:
 @$\begin{align*}a_n = 9(\frac{4}{3})^{(n1)}\end{align*}@$
 @$\begin{align*}a_n = 6(4)^{(n1)}\end{align*}@$
 @$\begin{align*}a_n = 5(5)^{(n1)}\end{align*}@$