Question

9What is the sum of the geometric series 500 (1.05)*¹m=1?

Answer 1

Given:

The geometric series

`\sum_{n\mathop{=}1}^9500(1.05)^(n-1)`

Required:

What is the sum of gemetric series?

Step-by-step explanation:

The formula for sum of geometric series with finite terms:

`\begin{gathered} S_n=(a(r^n-1))/(r-1) \\ Where, \\ a(first\text{ }term) \\ r(common\text{ }ratio) \end{gathered}`

So, the series:

`\begin{gathered} =500+500(1.05)+500(1.05)^2+...... \\ So,r=(500(1.05))/(500) \\ r=1.05 \end{gathered}`

Now, sum is:

`\begin{gathered} S_9=(500(1.05^9-1))/((9-1)) \\ =(500(0.551))/(8) \\ =(275.5)/(8) \\ =34.4 \end{gathered}`

Answer:

The sum is 34.4