Answer:
$68,000.
Sum of salaries in the first ten years:
[tex]\displaystyle \sum_{n = 1}^{10}{(\$50000 + (n - 1)\times \$2000)} = \$590000[/tex].
Step-by-step explanation:
Start by considering the salary for the first few years:
[tex]\begin{array}{c|c}\text{Year}&\text{Salary}\\\cline{1-2}\\[-0.5em]\text{1st}& \$50000+0\times \$2000\\\text{2nd}&\$50000 + 1\times\$2000\\\text{3rd}&\$50000 + 2\times \$2000\\\vdots&\vdots\\n\text{th}&\$50000 + (n - 1)\times \$2000\end{array}[/tex].
Observe the trend in annual salary. The salary on the nth year will be equal to [tex]\$50000 + (n - 1)\times \$2000[/tex]. The salary on the 10th year will thus be equal to
[tex]\$50000+(10-1)\times \$2000 = \$68000[/tex].
The sum of the salaries for the first ten years will be:
[tex]\displaystyle \sum_{n = 1}^{10}{(\$50000 + (n - 1)\times \$2000)}[/tex].
This series is arithmetic:
[tex]\begin{aligned}&\displaystyle \sum_{n = 1}^{10}{(\$50000 + (n - 1)\times \$2000)}\\=&\$50000 + \$52000 + \cdots + \$68000\end{aligned}[/tex].
Evaluate this expression using the formula for the sum of an arithmetic series:
[tex]\begin{aligned}S_n &= \frac{1}{2} n \cdot (a_1 + a_n)\\&= \frac{1}{2} \times 10 \times (\$68000 + \$50000) \\&= \$590000\end{aligned}[/tex]
