Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

If $|r| < 1$ , 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

$s_{n} = \frac{a_{1} (1 - r^{n})}{(1 - r)} s_{n} = \frac{a_{1}}{(1 - r)} - \frac{a_{1} r^{n})}{(1 - r)}$

If $|r| < 1$ then $|r^{n}|$ approaches $0$ as $n \to \infty$ then the term $\frac{a_{1} r^{n}}{(1 - r)}$ approaches $0$ and $s_{n}$ approaches $\frac{a_{1}}{(1 - r)}$ as $n \to \infty$ .

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

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Short Answer

If $|r| < 1$ , the sum of the geometric series a
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