# If a[1]=4 and r=2, what are the first four terms of the geometric sequence?

##### 1 Answer
Mar 13, 2016

$4 , 8 , 16 , 32$

#### Explanation:

Recall that the geometric sequence formula is written as:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{n} = a {r}^{n - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
${t}_{n} =$term number
$a =$first term
$r =$common ratio
$n =$number of terms

Determining the First Four Terms
$1$. Since the value of $a$ has already been given, the first term is $4$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{1} = 4 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , \ldots$

$2$. Using the geometric sequence formula, substitute your known values to determine the second term.

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

${t}_{2} = 4 {\left(2\right)}^{2 - 1}$

${t}_{2} = 4 {\left(2\right)}^{1}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{2} = 8 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , 8 , \ldots$

$3$. Repeat for the third term.

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

${t}_{3} = 4 {\left(2\right)}^{3 - 1}$

${t}_{3} = 4 {\left(2\right)}^{2}$

${t}_{3} = 4 \left(4\right)$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{3} = 16 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , 8 , 16 , \ldots$

$4$. Repeat for the fourth term.

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

${t}_{4} = 4 {\left(2\right)}^{4 - 1}$

${t}_{4} = 4 {\left(2\right)}^{3}$

${t}_{4} = 4 \left(8\right)$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{4} = 32 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , 8 , 16 , 32$

$\therefore$, the first four terms of the sequence are $4 , 8 , 16 , \text{and}$ $32$.