# In an arithmetic sequence, the first term is -2, the fourth term is 16, and the nth term is 11998, how do you find n and the common difference?

May 12, 2016

common difference, $d = 6$

$n = 2001$

#### Explanation:

${a}_{n} = a + \left(n - 1\right) d$

Here, $a$ is the first term, $d$ is the common difference and ${a}_{n}$ is the nth term of the sequence.

We are given:

• $a = - 2$

• ${a}_{4} = 16$

• ${a}_{n} = 11998$

$\textcolor{red}{\text{To find "n" and } d .}$

Let's start with finding $d$:

${a}_{n} = a + \left(n - 1\right) d$

$\textcolor{b r o w n}{\text{Put in "a=-2 " and } n = 4}$

${a}_{4} = - 2 + \left(4 - 1\right) d$

$16 = - 2 + \left(4 - 1\right) d$ , $\textcolor{b l u e}{\text{ since } {a}_{4} = 16}$

$16 = - 2 + 3 d$

Add $2$ to both sides:

$16 \textcolor{b l u e}{+ 2} = - 2 \textcolor{b l u e}{+ 2} + 3 d$

$18 = 3 d$

$\textcolor{red}{6 = d}$

Next, calculate $n$:

${a}_{n} = a + \left(n - 1\right) d$

$\textcolor{b r o w n}{\text{Put in "a_n=11998 " , " a=-2 " and } d = 6}$

$11998 = - 2 + \left(n - 1\right) 6$

Add $2$ to both sides:

$11998 \textcolor{b l u e}{+ 2} = - 2 \textcolor{b l u e}{+ 2} + \left(n - 1\right) 6$

$12000 = \left(n - 1\right) 6$

Divide both sides by $6$:

$\frac{12000}{\textcolor{b l u e}{6}} = \frac{\left(n - 1\right) 6}{\textcolor{b l u e}{6}}$

$2000 = n - 1$

Add $1$ to both sides:

$2000 \textcolor{b l u e}{+ 1} = n - 1 \textcolor{b l u e}{+ 1}$

$\textcolor{red}{2001 = n}$