Tommy spent six lucky days in Las Vegas. On his first day, he won a net amount of $20, and on each of the following days, the daily net amount he won grew by $$\$d$$. If Tommy won a total net amount of $1,620 during his stay in Las Vegas, how much did he win on the last day?

**Alternative Solution**:
Instead, you could solve for $$d$$ by writing out the amount Tommy won on each day and then writing an equation where they add up to $1,620.
- 1st day: $$\$20$$
- 2nd day: $$\$20 + \$d$$
- 3rd day: $$\$20 + \$2d$$
- 4th day: $$\$20 + \$3d$$
- 5th day: $$\$20 + \$4d$$
- 6th day: $$\$20 + \$5d$$
Adding those up and setting them equal to $1,620 gives us
> $$\$20\cdot 6 + \$15d = \$1{,}620$$
> $$\$120 + \$15d = \$1{,}620$$
> $$\$15d = \$1{,}500$$
> $$d = 100$$.
That means that on the sixth day, Tommy made $520.

Incorrect.
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Incorrect.
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Incorrect.
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Incorrect.
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Correct.
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Since the overall average amount he made is the average of the first and last days, it is
> $$\displaystyle \text{Average} = \frac{x+\$20}{2}$$.
Putting that into the formula for finding the average, you get
> $$\displaystyle \frac{x+\$20}{2} = \frac{\$ 1{,}620}{6}$$
> $$\displaystyle \frac{x+\$20}{2} = \$ 270$$
> $$\displaystyle x+\$20 = \$ 540$$
> $$\displaystyle x = \$ 520$$.
Hence, this is the correct answer.

$330

$500

$520

$540

$620

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