# What is the geometric sequence for 3, 6, 12, 24, ...?

Feb 10, 2017

The geometric sequence is ${a}_{n} = a {r}^{n - 1} = 3 \cdot {2}^{n - 1}$.

#### Explanation:

In a geometric sequence, the terms are separated by a common ratio $r$. So, for example, the 4th term ${a}_{4}$ will be $r \times {a}_{3}$, the 3rd term ${a}_{3} = r \times {a}_{2}$, and so on. From this we can get a general formula for the ${n}^{\text{th}}$ term in terms of $r$ and the first term ${a}_{1}$:

${a}_{n} = r \times {a}_{n - 1} \text{ } = {r}^{1} {a}_{n - 1}$
$\textcolor{w h i t e}{{a}_{n}} = r \times \left(r \times {a}_{n - 2}\right) \text{ } = {r}^{2} {a}_{n - 2}$
$\textcolor{w h i t e}{{a}_{n}} = r \times r \times \left(r \times {a}_{n - 3}\right) \text{ } = {r}^{3} {a}_{n - 3}$
$\textcolor{w h i t e}{{a}_{n}} = \ldots$
a_n=r^(n-1)xxa_(n-(n"-"1))"     "=r^(n-1)a_1

This is often written with the initial value ${a}_{1}$ (often just called $a$) in front, like this:

${a}_{n} = a {r}^{n - 1}$

For the sequence $3 , 6 , 12 , 24$, we are given the first term $a = 3$. Now all we need is the common ratio $r$. This can be found by computing the ratio of any two successive terms.

(That is, since ${a}_{2} = r \times {a}_{1}$ for any geometric sequence, we can find $r$ by solving $r = {a}_{2} / {a}_{1}$, or $r = {a}_{3} / {a}_{2}$, or in general $r = {a}_{k} / {a}_{k - 1}$.)

Using ${a}_{2} = 6$ and ${a}_{1} = 3$, we get

$r = {a}_{2} / {a}_{1} = \frac{6}{3} = 2$.

Thus, the common ratio is 2, the first term is 3, and so the formula for this geometric sequence is

${a}_{n} = a {r}^{n - 1}$
$\textcolor{w h i t e}{{a}_{n}} = 3 \cdot {2}^{n - 1}$.