Tom and Jerry each start a new job. Tom is paid $30,000 in the first year, and in every subsequent year, his annual pay will increase by $1,500. Jerry is paid $25,000 in the first year, and in every subsequent year, his annual pay will increase by 5%.
If they both start working at the beginning of 2013,

1. Express Tom's annual pay in the $n$th year

Let $T_n$ be Tom's pay in the $n$th year. Since his pay increases by a constant $1,500 each year, it forms an arithmetic progression with first term $a = 30000$ and common difference $d = 1500$.
$T_n = 30000 + 1500(n-1)$

2. Express Jerry's annual pay in the $n$th year

Let $J_n$ be Jerry's pay in the $n$th year. Since his pay increases by 5% each year, it forms a geometric progression with first term $a = 25000$ and common ratio $r = 1.05$.
$J_n = 25000(1.05)^{n-1}$

3. Formulate the inequality to find when Jerry's pay exceeds Tom's

We require $J_n > T_n$:
$25000(1.05)^{n-1} > 30000 + 1500(n-1)$

4. Solve the inequality using a Graphing Calculator (GC)

This mixed exponential-polynomial inequality cannot be solved algebraically. Using a graphing calculator to generate a table of values for $T_n$ and $J_n$:
When $n = 16$, $T_{16} = 52500$ and $J_{16} \approx 51973.11$, so $J_{16} < T_{16}$.
When $n = 17$, $T_{17} = 54000$ and $J_{17} \approx 54571.76$, so $J_{17} > T_{17}$.
The smallest integer $n$ is $17$.

5. Determine the actual calendar year

The 1st year ($n=1$) is 2013.
The 17th year is $2013 + 17 - 1 = 2029$.
Therefore, the first year Jerry will be paid more than Tom is 2029.

1. Formulate an expression for Tom's total income after $n$ years.

Tom's annual pay forms an arithmetic progression with first term $a = 30000$ and common difference $d = 1500$.
Let $S_T$ be his total income after $n$ years.
$S_T = \frac{n}{2}[2(30000) + (n-1)(1500)]$
$S_T = 750n^2 + 29250n$

2. Formulate an expression for Jerry's total income after $n$ years.

Jerry's annual pay forms a geometric progression with first term $a = 25000$ and common ratio $r = 1.05$.
Let $S_J$ be his total income after $n$ years.
$S_J = \frac{25000(1.05^n - 1)}{1.05 - 1}$
$S_J = 500000(1.05^n - 1)$

3. Set up the inequality and find the smallest integer $n$ using a graphing calculator.

We require Jerry's total income to exceed Tom's total income:
$S_J > S_T$
$500000(1.05^n - 1) > 750n^2 + 29250n$
Using the table function on a graphing calculator:
When $n = 25$, $S_T = 1200000$ and $S_J \approx 1193177$ (so $S_J < S_T$)
When $n = 26$, $S_T = 1267500$ and $S_J \approx 1277836$ (so $S_J > S_T$)

The inequality is first satisfied when $n = 26$.

4. Determine the corresponding calendar year.

The 1st year corresponds to 2013.
The 26th year corresponds to $2013 + 26 - 1 = 2038$.

Jerry's total income will first exceed Tom's at the end of 2038.