Chapter 10 - Section 10.2 - Arithmetic Sequences - Exercise Set - Page 1061: 68

$442,500$

We know that $a_n=a_1+(n-1) d$ Here, we have $a_1=33,000$ and $d=2500$ and $a_{10}=33, 000+2500(10-1)=55,500$ The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ $S_{10}=\dfrac{10}{2}[33,000+55,500]=442,500$