Question

I need help with this practice problem in my trigonometry prep book.

Answer 1

Sum of the First n Terms of a Geometric Sequence

Given a geometric sequence (or series) with a first-term a1 and common ratio r, the sum of the first n terms is given by:

$S_n=a_1\cdot\frac{1-r^n}{1-r}$

We are given the series:

$120-80+\frac{160}{3}-\frac{320}{9}+\cdots$

Before calculating the required sum, we need to find the common ratio. It's defined as the division of two consecutive terms. For example, using the first two terms:

$\begin{gathered} r=-\frac{80}{120} \\ Simplify\colon \\ r=-\frac{2}{3} \end{gathered}$

The first term is a1 =120. Now apply the formula:

$S_8=120\cdot\frac{1-(-\frac{2}{3})^8}{1+\frac{2}{3}}$

Operating:

$\begin{gathered} S_8=120\cdot\frac{1-\frac{2^8}{3^8}}{\frac{5}{3}} \\ S_8=120\cdot\frac{1-\frac{256}{6561}}{\frac{5}{3}} \\ S_8=120\cdot\frac{\frac{6561-256}{6561}}{\frac{5}{3}} \\ S_8=120\cdot\frac{\frac{6305}{6561}}{\frac{5}{3}} \end{gathered}$

Calculating:

$\begin{gathered} S_8=120\cdot\frac{1261}{2187} \\ \text{Simplifying:} \\ S_8=\frac{50440}{729} \end{gathered}$