Algebra Ii Worksheet 11.4 - Infinite Geometric Series & Word ... - Free Printable

Problem Analysis and Solution

The provided worksheet involves problems related to infinite geometric series and word problems. Let's solve each problem step by step.

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#### Part 1: Infinite Geometric Series

The general formula for the sum of an infinite geometric series is:
\[
S = \frac{a}{1 - r}
\]
where:
- \( a \) is the first term,
- \( r \) is the common ratio, and
- \( |r| < 1 \) for the series to converge.

If \( |r| \geq 1 \), the series does not have a finite sum.

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##### Problem 1: \( 24 + 12 + 6 + 3 + \cdots \)

- First term (\( a \)): \( 24 \)
- Common ratio (\( r \)): \( \frac{12}{24} = \frac{1}{2} \)

Since \( |r| = \frac{1}{2} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{24}{1 - \frac{1}{2}} = \frac{24}{\frac{1}{2}} = 24 \times 2 = 48
\]

Answer: \( 48 \)

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##### Problem 2: \( 24 - 12 + 6 - 3 + \cdots \)

- First term (\( a \)): \( 24 \)
- Common ratio (\( r \)): \( \frac{-12}{24} = -\frac{1}{2} \)

Since \( |r| = \left|-\frac{1}{2}\right| = \frac{1}{2} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{24}{1 - \left(-\frac{1}{2}\right)} = \frac{24}{1 + \frac{1}{2}} = \frac{24}{\frac{3}{2}} = 24 \times \frac{2}{3} = 16
\]

Answer: \( 16 \)

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##### Problem 3: \( 27 - 18 + 12 - 8 + \cdots \)

- First term (\( a \)): \( 27 \)
- Common ratio (\( r \)): \( \frac{-18}{27} = -\frac{2}{3} \)

Since \( |r| = \left|-\frac{2}{3}\right| = \frac{2}{3} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{27}{1 - \left(-\frac{2}{3}\right)} = \frac{27}{1 + \frac{2}{3}} = \frac{27}{\frac{5}{3}} = 27 \times \frac{3}{5} = \frac{81}{5} = 16.2
\]

Answer: \( 16.2 \) or \( 16 \frac{1}{5} \)

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##### Problem 4: \( 27 + 18 + 12 + 8 + \cdots \)

- First term (\( a \)): \( 27 \)
- Common ratio (\( r \)): \( \frac{18}{27} = \frac{2}{3} \)

Since \( |r| = \frac{2}{3} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 27 \times 3 = 81
\]

Answer: \( 81 \)

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##### Problem 5: \( 256 + 320 + 400 + 500 + \cdots \)

- First term (\( a \)): \( 256 \)
- Common ratio (\( r \)): \( \frac{320}{256} = \frac{5}{4} \)

Since \( |r| = \frac{5}{4} > 1 \), the series does not converge. Therefore, it has no sum.

Answer: No sum

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##### Problem 6: \( 500 + 400 + 320 + 256 + \cdots \)

- First term (\( a \)): \( 500 \)
- Common ratio (\( r \)): \( \frac{400}{500} = \frac{4}{5} \)

Since \( |r| = \frac{4}{5} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{500}{1 - \frac{4}{5}} = \frac{500}{\frac{1}{5}} = 500 \times 5 = 2500
\]

Answer: \( 2500 \)

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##### Problem 7: \( 3 + 4 + 5 \frac{1}{3} + 7 \frac{1}{9} + \cdots \)

This series is not geometric because the ratio between consecutive terms is not constant. For example:
\[
\frac{4}{3} \neq \frac{5 \frac{1}{3}}{4}
\]
Thus, it does not fit the definition of a geometric series, and we cannot use the formula for the sum of a geometric series.

Answer: No sum

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##### Problem 8: \( \frac{1}{2} - \frac{1}{3} + \frac{2}{9} - \frac{4}{27} + \cdots \)

- First term (\( a \)): \( \frac{1}{2} \)
- Common ratio (\( r \)): \( \frac{-\frac{1}{3}}{\frac{1}{2}} = -\frac{2}{3} \)

Since \( |r| = \left|-\frac{2}{3}\right| = \frac{2}{3} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \left(-\frac{2}{3}\right)} = \frac{\frac{1}{2}}{1 + \frac{2}{3}} = \frac{\frac{1}{2}}{\frac{5}{3}} = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10}
\]

Answer: \( \frac{3}{10} \) or \( 0.3 \)

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##### Problem 9: \( \sum_{n=0}^{\infty} 3 \left(\frac{1}{4}\right)^n \)

- First term (\( a \)): \( 3 \left(\frac{1}{4}\right)^0 = 3 \)
- Common ratio (\( r \)): \( \frac{1}{4} \)

Since \( |r| = \frac{1}{4} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 3 \times \frac{4}{3} = 4
\]

Answer: \( 4 \)

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##### Problem 10: \( \sum_{n=1}^{\infty} \frac{2^n}{5^n} \)

- First term (\( a \)): \( \frac{2^1}{5^1} = \frac{2}{5} \)
- Common ratio (\( r \)): \( \frac{\frac{2^{n+1}}{5^{n+1}}}{\frac{2^n}{5^n}} = \frac{2}{5} \)

Since \( |r| = \frac{2}{5} < 1 \), the series converges. Using the formula:
\[
S = \frac{a}{1 - r} = \frac{\frac{2}{5}}{1 - \frac{2}{5}} = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{5} \times \frac{5}{3} = \frac{2}{3}
\]

Answer: \( \frac{2}{3} \)

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#### Part 2: Word Problems

##### Problem 11: Clock Chimes

The clock chimes as many times as the hour. From 1 A.M. to midnight (12 P.M.), the hours are:
\[
1, 2, 3, \ldots, 12
\]
This is an arithmetic series with:
- First term (\( a \)): \( 1 \)
- Last term (\( l \)): \( 12 \)
- Number of terms (\( n \)): \( 12 \)

The sum of an arithmetic series is given by:
\[
S = \frac{n}{2} (a + l)
\]
Substituting the values:
\[
S = \frac{12}{2} (1 + 12) = 6 \times 13 = 78
\]

However, the problem asks for the total number of chimes from 1 A.M. to midnight, which includes both A.M. and P.M. hours. Thus, we need to double the sum:
\[
\text{Total chimes} = 78 \times 2 = 156
\]

Answer: \( 156 \) [Arithmetic]

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##### Problem 12: Theater Seats

The front row has 25 seats, and each subsequent row has 2 more seats than the previous row. This forms an arithmetic sequence where:
- First term (\( a \)): \( 25 \)
- Common difference (\( d \)): \( 2 \)
- Number of rows (\( n \)): \( 20 \)

The total number of seats in the first 20 rows is the sum of the first 20 terms of this arithmetic sequence. The sum of an arithmetic series is given by:
\[
S_n = \frac{n}{2} [2a + (n - 1)d]
\]
Substituting the values:
\[
S_{20} = \frac{20}{2} [2(25) + (20 - 1)(2)] = 10 [50 + 19 \cdot 2] = 10 [50 + 38] = 10 \times 88 = 880
\]

Answer: \( 880 \) seats [Arithmetic]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & 48 \\
2. & 16 \\
3. & 16.2 \text{ or } 16 \frac{1}{5} \\
4. & 81 \\
5. & \text{No sum} \\
6. & 2500 \\
7. & \text{No sum} \\
8. & \frac{3}{10} \text{ or } 0.3 \\
9. & 4 \\
10. & \frac{2}{3} \\
11. & 156 \\
12. & 880 \\
\end{array}
}
\]