Precalculus Examples

$120 + 40 + \frac{40}{3} + \frac{40}{9}$

Step 1

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by $\frac{1}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{3}$

Step 2

The sum of a series $S_{n}$ is calculated using the formula $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ . For the sum of an infinite geometric series $S_{\infty}$ , as $n$ approaches $\infty$ , $1 - r^{n}$ approaches $1$ . Thus, $\frac{a (1 - r^{n})}{1 - r}$ approaches $\frac{a}{1 - r}$ .

$S_{\infty} = \frac{a}{1 - r}$

Step 3

The values $a = 120$ and $r = \frac{1}{3}$ can be put in the equation $S_{\infty}$ .

$S_{\infty} = \frac{120}{1 - \frac{1}{3}}$

Step 4

Simplify the equation to find $S_{\infty}$ .

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Step 4.1

Simplify the denominator.

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Step 4.1.1

Write $1$ as a fraction with a common denominator.

$S_{\infty} = \frac{120}{\frac{3}{3} - \frac{1}{3}}$

Step 4.1.2

Combine the numerators over the common denominator.

$S_{\infty} = \frac{120}{\frac{3 - 1}{3}}$

Step 4.1.3

Subtract $1$ from $3$ .

$S_{\infty} = \frac{120}{\frac{2}{3}}$

Step 4.2

Multiply the numerator by the reciprocal of the denominator.

$S_{\infty} = 120 (\frac{3}{2})$

Step 4.3

Cancel the common factor of $2$ .

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Step 4.3.1

Factor $2$ out of $120$ .

$S_{\infty} = 2 (60) (\frac{3}{2})$

Step 4.3.2

Cancel the common factor.

$S_{\infty} = 2 \cdot (60 (\frac{3}{2}))$

Step 4.3.3

Rewrite the expression.

$S_{\infty} = 60 \cdot 3$

Step 4.4

Multiply $60$ by $3$ .

$S_{\infty} = 180$