Math Help for Section 13.2, Page 677

Example
8: Tip
Instead of listing the sales for each year to find the sales in the
tenth year, \$325,000, you could have used $a_1=\$100,000$ and
$d=\$25,000$
to calculate

    $\eqalign{a_n=&a_1+(n-1)d
&{\small\color{red}\text{Formula for }n\text{th term of an
arithmetic sequence}}  \cr =&
{\color{red}100,000}+{\color{red}25,000}(n-1) 
&{\small\color{red}\text{Substitute 100,000 for
}a_1\text{
and 25,000 for }d.}\cr
=&100,000+25,000n-25,000
\quad\quad &{\small\color{red}\text{Distributive
Property}} \cr =& 25,000n+75,000.
&{\small\color{red}\text{Combine like terms.}} }
$

So, $a_{10}=25,000({\color{red}10})+75,000=\$325,000$.

Or you could have realized that the sales are
\$100,000 in the first year and will increase by \$25,000 nine times
between the first and tenth years, so the sales for the tenth year are

    $a_{10}=100,000+9(25,000)=\$325,000$.

Then proceed with the $n\text{th}$ partial sum formula as shown on page 589.