A. lottery plans to give out 5,000,000 in prizes.The first ticket drawn wins 20 the second ticket drawn wins 50 the third ticket drawn wins 125 and so on. Can the lottery afford to give out 14 prizes? y=1.5 5,000,000=

## Step 1: Calculate the common ratio<br />### The prize money increases by a factor of 2.5 each time. $50/20 = 2.5$ and $125/50 = 2.5$. This is a geometric sequence.<br /><br />## Step 2: Calculate the total prize money for 14 prizes<br />### The sum of a geometric series is given by $S_n = \frac{a_1(r^n - 1)}{r-1}$, where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. In this case, $a_1 = 20$, $r = 2.5$, and $n = 14$. So, $S_{14} = \frac{20(2.5^{14} - 1)}{2.5 - 1} = \frac{20(2.5^{14} - 1)}{1.5}$.<br /><br />## Step 3: Evaluate the sum<br />### $S_{14} = \frac{20(372529.0298 - 1)}{1.5} \approx \frac{20(372528.0298)}{1.5} \approx \frac{7450560.596}{1.5} \approx 4967040.397$.<br /><br />## Step 4: Compare with the available budget<br />### The total prize money for 14 prizes is approximately $\$4,967,040.40$, which is less than the $\$5,000,000$ budget.