# Geometric Sequences and Exponential Functions

## Key Questions

• A geometric sequence is given by a starting number, and a common ratio.

Each number of the sequence is given by multipling the previous one for the common ratio.

Let's say that your starting point is $2$, and the common ratio is $3$. This means that the first number of the sequence, ${a}_{0}$, is 2. The next one, ${a}_{1}$, will be $2 \setminus \times 3 = 6$. In general, we have that ${a}_{n} = 3 {a}_{n - 1}$.

If the starting point is $a$, and the ratio is $r$, we have that the generic element is given by ${a}_{n} = a {r}^{n}$. This means that we have several cases:

1. If $r = 1$, the sequence is constantly equal to $a$;
2. If $r = - 1$, the sequence is alternatively equal to $a$ and $- a$;
3. If $r > 1$, the sequence grows exponentially to infinity;
4. If $r < - 1$, the sequence grows to infinity, assuming alternatively positive and negative values;
5. If $- 1 < r < 1$, the sequence exponentially decreases to zero;
6. If $r = 0$, the sequence is constantly zero, from the second term on.
• It means that the next term can always be obtained by multiplying a current term by $r$.

${a}_{1} = a$

by multiplying by $r$,

${a}_{2} = a r$

by multiplying by $r$,

${a}_{3} = a {r}^{2}$
.
.
.

I hope that this was helpful.

• A geometric sequence is always of the form
${t}_{n} = {t}_{\text{n-1}} \cdot r$

Every next term is $r$ times as large as the one before.

So starting with ${t}_{0}$ (the "start term") we get:
${t}_{1} = r \cdot {t}_{0}$
${t}_{2} = r \cdot {t}_{1} = r \cdot r \cdot {t}_{0} = {r}^{2} \cdot {t}_{0}$
......
${t}_{n} = {r}^{n} \cdot {t}_{0}$

${t}_{n} = {r}^{n} \cdot {t}_{0}$
${t}_{0}$ being the start term, $r$ being the ratio

Extra:
If $r > 1$ then the sequence is said to be increasing
if $r = 1$ then all numbers in the sequence are the same
If $r < 1$ then the sequence is said to be decreasing ,
and a total sum may be calculated for an infinite sequence:
sum $\sum = {t}_{0} / \left(1 - r\right)$

Example :
The sequence $1 , \frac{1}{2} , \frac{1}{4} , \frac{1}{8.} . .$
Here the ${t}_{0} = 1$ and the ratio $r = \frac{1}{2}$
Total sum of this infinite sequence:
$\sum = {t}_{0} / \left(1 - r\right) = \frac{1}{1 - \frac{1}{2}} = 2$

• This key question hasn't been answered yet.

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