50 points!!! What is the sum of the first 8 terms of the geometric series? 120 - 80 + 160/3 - 320/9 + ... Express your answer as a simplified fraction. Show your work please!!!

To find the sum of the first 8 terms of a geometric series, we first need to determine the common ratio (r) of the series.

Looking at the given terms:

1st term, a = 120 2nd term = -80 3rd term = 160/3

To find the common ratio, we divide the second term by the first term:

r = (-80) / 120 = -2/3

Now, to verify if this is indeed a geometric series with common ratio -2/3, we check the third term:

r = (160/3) / (-80) = (-2/3)

Since the ratio is consistent, we can confirm that the common ratio r is indeed -2/3.

The sum of the first n terms of a geometric series can be found using the formula:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

where, - \( S_n \) is the sum of the first n terms, - \( a \) is the first term, and - \( r \) is the common ratio.

Using this formula, we calculate the sum of the first 8 terms, where n = 8:

\[ S_8 = 120 \frac{1 - (-2/3)^8}{1 - (-2/3)} \]

Now let's simplify this step by step:

First, let's find \( (-2/3)^8 \): \[ (-2/3)^8 = \frac{(-2)^8}{3^8} = \frac{256}{6561} \]

Then rewrite the sum formula:

\[ S_8 = 120 \frac{1 - \frac{256}{6561}}{1 - (-2/3)} \]

\[ S_8 = 120 \frac{\frac{6561 - 256}{6561}}{\frac{2 + 3}{3}} \]

\[ S_8 = 120 \frac{6305}{6561} \cdot \frac{3}{5} \]

Now perform the multiplications:

\[ S_8 = \frac{120 \cdot 6305 \cdot 3}{6561 \cdot 5} \]

\[ S_8 = \frac{360 \cdot 6305}{6561} \]

Finally, simplifying this fraction, we get the sum:

\[ S_8 = \frac{2269800}{6561} \]

This fraction does not appear to be reducible further, so this is the simplified fraction representing the sum of the first 8 terms of the given geometric series.