Answer
8 years
Work Step by Step
The salaries form an arithmetic sequence, with
with $a_{1}=35,000,d=1400,$
We want n for which $S_{n}=280,000$.
$S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$ ... which we solve for n
\begin{aligned}
280,000&=\displaystyle \frac{n}{2}[2(35,000)+(n-1)1400]\\
280,000&=\displaystyle \frac{n}{2}\cdot2[35,000+700(n-1)]\\
280,000&=n[35,000+700n-700]\\
280,000&=n(700n+34,300)\\
280,000&=700n^{2}+34,300n\\
400&=n^{2}+49n\\
n^{2}+49n-400&=0\end{aligned}
...use the quadratic formula
$n=\displaystyle \frac{-49\pm\sqrt{49^{2}-4(1)(-400)}}{2(1)}\approx\frac{-49\pm 63.25}{2}$
... discard the negative solution
$n\approx 7.13$
We are discussing end-of-year salaries.
This result means that 7 years is not enough.
It takes 8 years for the aggregate salary to reach and be above ${{\$}} 280,000$.
