College Algebra, Chapter 9, 9.2, Section 9.2, Problem 38

Suppose the 12th term of an arithmetic sequence is $32$, and the fifth term is $18$. Find the 20th term.

To find the $n$th term of this sequence, we need to find $a$ and $d$ in the formula

$a_n = a + (n -1) d$

From this formula we get

$a_{12} = a + (12-1) d = a+ 11d$

$a_5 = a + (5-1)d = a + 4d$

Since $a_5 = 18$ and $a_{12} = 32$, we get the two equations



$
\left\{
\begin{equation}
\begin{aligned}

a + 11d =& 32
&& \text{Equation 1}
\\
\\
a + 4d =& 18
&& \text{Equation 2}

\end{aligned}
\end{equation}
\right.
$


We eliminate $a$-term in each equations and solve for $d$



$
\left\{
\begin{equation}
\begin{aligned}

a + 11d =& 32 &&
\\
\\
-a-4d =& -18
&& -1 \times \text{Equation 2}
\\
\\
\end{aligned}
\end{equation}
\right.
$


$
\qquad
\begin{equation}
\begin{aligned}

\hline\\
\\
\\
7d =& 14
&& \text{Add}
\\
\\
d =& \frac{14}{7}
&& \text{Divide by } 7
\\
\\
d =& 2
&&

\end{aligned}
\end{equation}
$


We back-substitute $d =2$ into the first equation and solve for $a$


$
\begin{equation}
\begin{aligned}

a+11(2) =& 32
&& \text{Back-substitute } d=2
\\
\\
a =& 32-22
&& \text{Subtract } 11(2)=22
\\
\\
a =& 10
&&

\end{aligned}
\end{equation}
$


So we get $a=10$ and $d=2$. Thus, the $n$th term of this sequence is $a_n = 10+2(n-1)$

The 20th term is

$a_{20} = 10 + 2(20 - 1) = 48$