#### Answer

The salary for the sixth year will be\[\$38,288\].

#### Work Step by Step

In the first year, the salary of the player is $30,000 and annual year the increment is 5%.
So,
\[\begin{align}
& 5%\text{ of }30000=\frac{5}{100}\times 30000 \\
& =5\times 300 \\
& =1500
\end{align}\]
So, the salary of the player at the starting of the second year is:
\[\$30,000+\$1,500=\$31,500\]
At the starting of the third year, the salary of the player will be again increased by 5%.
\[\begin{align}
& 5%\text{ of }31500=\frac{4}{100}\times 31500 \\
& =4\times 315 \\
& =1260
\end{align}\]
The salary for the third year is:
\[\$31,500+\$1,260=\$32,760\]
Now the series is 30000, 31500, 32760……. It is a form of G.P
The nth term is found in G.P with the help of the following formula
\[{{a}_{n}}=a{{r}^{\left( n-1 \right)}}\]
The salary of the player accumulates according to G.P. with \[a=30000,\,r=1\cdot 05\] and it is required to find \[{{6}^{th}}\]term.
\[\begin{align}
& {{a}_{n}}=a{{r}^{n-1}} \\
& {{a}_{6}}=30000\times {{\left( 1.05 \right)}^{\left( 6-1 \right)}} \\
& =30000\times {{\left( 1.05 \right)}^{5}} \\
& {{a}_{6}}=38288.44
\end{align}\]
Hence, the salary in year 6 rounded off to the nearest dollar will be\[\$38,288\].