Question

find the sum of the first 6 terms of the following geometric sequence: 50, 40, 32, \frac{128}{5}, \dots hint: $s = \frac{a(1 - r^n)}{1 - r}$ round your answer to the nearest hundredth.

Explanation:

Step1: Identify \(a\), \(r\), and \(n\)

In a geometric sequence, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. Here, \(a = 50\), \(n = 6\). To find \(r\), divide the second term by the first term: \(r=\frac{40}{50}=\frac{4}{5} = 0.8\).

Step2: Apply the sum formula

The formula for the sum of the first \(n\) terms of a geometric sequence is \(S=\frac{a(1 - r^{n})}{1 - r}\). Substitute \(a = 50\), \(r = 0.8\), and \(n = 6\) into the formula:
\[

$$\begin{align*} S&=\frac{50(1-(0.8)^{6})}{1 - 0.8}\\ &=\frac{50(1 - 0.262144)}{0.2}\\ &=\frac{50\times0.737856}{0.2}\\ &=\frac{36.8928}{0.2}\\ &= 184.464 \end{align*}$$

\]

Step3: Round to the nearest hundredth

Rounding \(184.464\) to the nearest hundredth gives \(184.46\).

Answer:

\(184.46\)