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Q 2.

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Found in: Page 824

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# If $\left|r\right|<1$ , the sum of the geometric series a$\sum _{k=1}^{\infty }a{r}^{k-1}$ is ________ .

If$\left|r\right|<1$ , the sum of the geometric series a $\sum _{k=1}^{\infty }a{r}^{k-1}$is $\frac{{a}_{1}}{\left(1-r\right)}$.

See the step by step solution

## Step 1. Given information.

If $\left|r\right|<1$, the sum of the geometric series a $\sum _{k=1}^{\infty }a{r}^{k-1}$ is _________.

• A geometric progression, also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

## Step 2. Fill in the blanks.

The sum ${s}_{n}$ of first $n$ terms of a geometric sequence is given by

${s}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{\left(1-r\right)}\phantom{\rule{0ex}{0ex}}{s}_{n}=\frac{{a}_{1}}{\left(1-r\right)}-\frac{{a}_{1}{r}^{n}\right)}{\left(1-r\right)}$

If $\left|r\right|<1$ then $\left|{r}^{n}\right|$ approaches $0$ as$n\to \infty$ then the term$\frac{{a}_{1}{r}^{n}}{\left(1-r\right)}$ approaches $0$ and ${s}_{n}$ approaches $\frac{{a}_{1}}{\left(1-r\right)}$as $n\to \infty$.

Therefore, if $\left|r\right|<1$the sum of the geometric series a $\sum _{k=1}^{\infty }a{r}^{k-1}$ is $\frac{{a}_{1}}{\left(1-r\right)}$.