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Algebra II : Geometric Sequences

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Example Question #21 : Geometric Sequences

A sequence begins as follows:

$\displaystyle 128, -64, 32, -16, ...$

Which statement is true?

Possible Answers:

The sequence may be geometric.

None of these

The sequence may be arithmetic and geometric.

The sequence may be arithmetic.

The sequence cannot be arithmetic or geometric.

Correct answer:

The sequence may be geometric.

Explanation:

An arithmetic sequence is one in which each term is generated by adding the same number - the common difference - to the previous term. As can be seen here, the difference between each term and the previous term varies from term to term:

$\displaystyle -64 - 128 = -192$

$\displaystyle 32- 64 = -32$

The sequence cannot be arithmetic.

A geometric sequence is one in which each term is generated by multiplying the previous term by the same number - the common ratio. As can be seen here, the ratio of each term to the previous one is the same:

$\frac{-64}{128} = -\frac{1}{2}$ $\displaystyle \frac{-64}{128} = -\frac{1}{2}$

$\frac{32}{-64} = -\frac{1}{2}$ $\displaystyle \frac{32}{-64} = -\frac{1}{2}$

$\frac{-16}{32} = -\frac{1}{2}$ $\displaystyle \frac{-16}{32} = -\frac{1}{2}$

The sequence could be geometric.

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Example Question #22 : Geometric Sequences

Find the sum for the first 25 terms in the series $60 + 40 + 26. \overline{6} + 17. \overline{7 } + ...$ $\displaystyle 60 + 40 + 26. \overline{6} + 17. \overline{7 } + ...$

Possible Answers:

19.99 $\displaystyle 19.99$

59.33 $\displaystyle 59.33$

$\displaystyle 30$

179.99 $\displaystyle 179.99$

$\displaystyle 3,030, 020. 19$

Correct answer:

179.99 $\displaystyle 179.99$

Explanation:

Before we add together the first 25 terms, we need to determine the structure of the series. We know the first term is 60. We can find the common ratio r by dividing the second term by the first:

$r = 40 \div 60 = \frac{2}{3}$ $\displaystyle r = 40 \div 60 = \frac{2}{3}$

We can use the formula $\sum_{k=0}^{n} Ar^k = A* \frac{1-r^n}{1-r}$ $\displaystyle \sum_{k=0}^{n} Ar^k = A* \frac{1-r^n}{1-r}$ where A is the first term.

The terms we are adding together are $60 *( \frac{2}{3} )^ 0 + 60 * (\frac {2}{3} ) ^ 1 + ... + 60 * (\frac{2}{3})^{24 }$ $\displaystyle 60 *( \frac{2}{3} )^ 0 + 60 * (\frac {2}{3} ) ^ 1 + ... + 60 * (\frac{2}{3})^{24 }$ so we can plug in $A = 60, r = \frac{2}{3 } , n = 24$ $\displaystyle A = 60, r = \frac{2}{3 } , n = 24$ :

$60 * \frac{1 -( \frac{2}{3})^{24}}{1 - \frac{2}{3}} \approx 60 * \frac{1 - 0.000059 }{ \frac{1}{3}} \approx 179.99$ $\displaystyle 60 * \frac{1 -( \frac{2}{3})^{24}}{1 - \frac{2}{3}} \approx 60 * \frac{1 - 0.000059 }{ \frac{1}{3}} \approx 179.99$

Common mistakes would involve order of operations - make sure you do exponents first, then subtract, then multiply/divide based on what is grouped together.

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Algebra II : Geometric Sequences

Study concepts, example questions & explanations for Algebra II

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Example Question #21 : Geometric Sequences

Example Question #22 : Geometric Sequences

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