Answer
$a_{n}=5n-9$
$a_{20}=91$
Work Step by Step
If the sequence is arithmetic, the general rule is
$ a_{n}=a_{1}+(n-1)d\qquad$
(Our goal is to find $a_{1}$ and $d.)$
Use the given information
$\left\{\begin{array}{lll}
26=a_{1}+(7-1)d & ... & (n=7)\\
71=a_{1}+(16-1)d & ... & (n=6)
\end{array}\right.\qquad$... simplify both equations
$\left\{\begin{array}{ll}
26=a_{1}+6d & \\
71=a_{1}+15d &
\end{array}\right.\qquad$... subtract the first from the second equation
$ 45=9d \qquad$... divide with 9
$5=d$
... back-substitute into $26=a_{1}+6d$
$26=a_{1}+6(5)$
$26=a_{1}+30$
$-4=a_{1}$
So, the general rule is
$a_{n}=-4+(n-1)(5)$
$a_{n}=-4+5n-9$
$a_{n}=5n-9$
The 20th term is
$a_{20}=5(20)-9=91$
