Geometric Sequences - Algebra II

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Question

$\textup{In the above geometric sequence, what is the value of }n\textup{?}$

Answer

$\textup{Look for a pattern. Each term in the sequence is three times the previous term.}$

$18\ast3=54$

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Question

Which of the following is a geometric sequence?

Answer

A geometric sequence is one in which the next term is found by mutlplying the previous term by a particular constant. Thus, we look for an implicit definition which involves multiplication of the previous term. The only possibility is:

$t_{0}=4$

$t_{n}=t_{n-1}\times 3$

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Question

$\left { -1, 2, -4, 8, -16, 32... \right }$

What is the explicit formula for the above sequence? What is the 20th value?

Answer

This is a geometric series. The explicit formula for any geometric series is:

$a_{n}= a_{1} \cdot r^{n-1}$ , where is the common ratio and is the number of terms.

In this instance and .

$a_{n}= a_{1} \cdot r^{n-1}\rightarrow a_n=(-1)\cdot(-2)^{n-1}$

Substitute into the equation to find the 20th term:

$a_{20}=(-1)\cdot(-2)^{20-1}$

$a_{20}=-(-2)^{19}=-(-524288)=524288$

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Question

What type of sequence is shown below?

$\left { 4,9,15,22,30... \right }$

Answer

This series is neither geometric nor arithmetic.

A geometric sequences is multiplied by a common ratio () each term. An arithmetic series adds the same additional amount () to each term. This series does neither.

Mutiplicative and subtractive are not types of sequences.

Therefore, the answer is none of the other answers.

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Question

Identify the 10th term in the series:

$\left { \frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32}... \right }$

Answer

The explicit formula for a geometric series is $a_{n}=a_{1}\cdot r^{n-1}$

In this problem $r=\frac{1}{2}$

Therefore:

$a_{10}=\frac{1}{4}\cdot \frac{1}{2}^{9}$

${\color{Red} a_{10}=\frac{1}{2048}}$

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Question

Find the 15th term of the following series:

$\left { 120, 60, 30, 15, 7.5,... \right }$

Answer

This series is geometric. The explicit formula for any geometric series is:

$a_n=a_1\cdot r^{n-1}$

Where represents the $n^{th}$ term, is the first term, and is the common ratio.

In this series $r = \frac{1}{2}$ .

Therefore the formula to find the 15th term is:

$a_{15}=120\cdot \left(\frac{1}{2}\right)^{14}$

$a_{15}=.0073$

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Question

Which of the following could be the formula for a geometric sequence?

Answer

The explicit formula for a geometric series is $a_n=a_1\cdot r^{n-1}$ .

Therefore, ${\color{Red} a_6=3\cdot 7^5, a_1=3}$ is the only answer that works.

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Question

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

Answer

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:

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

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

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

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

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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Question

Give the 33rd term of the Geometric Series

\[2 is the first term\]

Answer

First we need to find the common ratio by dividing the second term by the first:

$-3 \div 2 = -1.5$

The $n^{th}$ term is

$\a_{n} = a_1(r)^{n-1}\a_n=2(-1.5)^{n-1}$ ,

so the 33rd term will be

$a_{33} = 2(-1.5)^ {32 } = 862,879.767$ .

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Question

Find the 19th term of the sequence $7,000, \enspace 2,800, \enspace 1,120, \enspace 448, ...$

\[the first term is 7,000\]

Answer

First find the common ratio by dividing the second term by the first:

$2,800 \div 7,000 = \frac{ 2}{5} = 0.4$

Since the first term is , the nth term can be found using the formula

$\a_{n} = a_1(r)^{n-1}$

$7,000(0.4)^ {n-1 }$ ,

so the 19th term is $7,000(0.4)^{18} = 0.000481$

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Question

Find the 21st term of the sequence

\[90 is 1st, so n=1\]

Answer

First, find the common ratio by dividing the second term by the first:

$108 \div 90 = 1.2$

The nth term can be found using

$\a_{n} = a_1(r)^{n-1}$

$90(1.2)^ {n-1}$ ,

so the 21st term is

$90(1.2) ^ {20 } = 3,450.384$ .

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Question

Find the 26th term of the sequence $1,771,561, \enspace -1,449,459, \enspace 1,185,921, \enspace -970,299, ...$

Answer

First we need to find the common ratio, which we can do by dividing the second term by the first:

$-1,449,459 \div 1,771,561 = 0.\overline{81}$

The first term is $1,771, 561 (-0.\overline{81} ) ^ 0$ , the second term is $1,771, 561 (-0.\overline{81})^ 1$ , so the 26th term is $1,771, 561 (-0.\overline{81})^{ 25} = -11,738.206$

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Question

Find the common ratio for this geometric series:

Answer

Find the common ratio for this geometric series:

The common ratio of a geometric series can be found by dividing any term by the term before it. More generally:

$r=\frac{N_t}{N_{t-1}}$

So, do try the following:

$r=\frac{343}{49}=7$

So our common ratio is 7

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Question

Find the next term in this geometric series:

Answer

Find the next term in this geometric series:

to find the next term, we must first find the common ratio.

The common ratio of a geometric series can be found by dividing any term by the term before it. More generally:

$r=\frac{N_t}{N_{t-1}}$

So, do try the following:

$r=\frac{343}{49}=7$

So our common ratio is 7

Next, find the next term in the series by multiplying our last term by 7

$N_{t+1}=2401*7=16807$

Making our next term 16807

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Question

What is if $S_2 = \frac{2}{3}$ and $r=\frac{1}{6}$ ?

Answer

Use the geometric series summation formula.

$\sum_{i=1}^{n}a_i=a(\frac{1-r^n}{1-r})$

Substitute $S_2 = \frac{2}{3}$ into $\sum_{i=1}^{n}a_i$ , and replace the and terms. The value of is two.

$\frac{2}{3}=a(\frac{1-(\frac{1}{6})^2}{1-\frac{1}{6}})$

Simplify the terms on the right and solve for .

$\frac{2}{3}=a(\frac{1-\frac{1}{36}}{1-\frac{1}{6}})$

$\frac{2}{3}=a(\frac{\frac{35}{36}}{ \frac{5}{6}} )$

Rewrite the complex fraction using a division sign.

$\frac{2}{3}=a(\frac{35}{36} \div \frac{5}{6})$

Change the sign from division to a multiplication sign and switch the second term.

$\frac{2}{3}=a(\frac{35}{36} \times \frac{6}{5})$

Simplify the terms inside the parentheses.

$\frac{2}{3}=a(\frac{7}{6})$

Isolate the variable by multiplying six-seventh on both sides.

$\frac{2}{3}\cdot \frac{6}{7}=a(\frac{7}{6}) \cdot \frac{6}{7}$

Simplify both sides.

$a=\frac{4}{7}$

The answer is: $\frac{4}{7}$

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Question

What is the next term given the following terms? $[\frac{1}{3},\frac{4}{3}, \frac{16}{3},...]$

Answer

Divide the second term with the first term, and the third term with the second term.

$\frac{4}{3}\div\frac{1}{3} = \frac{4}{3}\times \frac{3}{1} = 4$

$\frac{16}{3}\div\frac{4}{3}=\frac{16}{3}\times\frac{3}{4}=4$

The common ratio of this geometric sequence is four.

Multiply the third term by four.

$\frac{16}{3}\times 4 = \frac{64}{3}$

The answer is: $\frac{64}{3}$

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Question

Determine the 10th term if the first term is $\frac{1}{3}$ and the common ratio is $\frac{2}{5}$ . Answer in scientific notation.

Answer

Write the formula to find the n-th term for a geometric sequence.

$a_n = a_1\cdot r^{n-1}$

Substitute the known values into the equation.

$a_{10} = (\frac{1}{3})\cdot (\frac{2}{5})^{10-1}= \frac{512}{5859375}$

This fraction is equivalent to: $8.738\times 10^{-5}$

The 19th term is: $8.738\times 10^{-5}$

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Question

Determine the sum: $10+1+\frac{1}{10}+...+\frac{1}{10,000}$

Answer

Write the sum formula for a geometric series.

$\textup{Sum} = \frac{a_1(1-r^n)}{1-r}$

$a_1 =\textup{ first term} = 10$

$r=\textup{common ratio} = \frac{1}{10}$

Identify the number of terms that exist.

$[10,1,\frac{1}{10},\frac{1}{100},\frac{1}{1,000},\frac{1}{10,000}]$

There are six existing terms:

Substitute the terms into the formula.

$\textup{Sum} = \frac{a_1(1-r^n)}{1-r} = \frac{10(1-(\frac{1}{10})^6)}{1-\frac{1}{10}} = \frac{10-\frac{1}{100,000}}{\frac{10}{10}-\frac{1}{10}}$

$= \frac{\frac{1,000,000}{100,000}-\frac{1}{100,000}}{\frac{10}{10}-\frac{1}{10}} = \frac{\frac{999,999}{100,000}}{\frac{9}{10}}$

Simplify the complex fraction.

$\frac{999,999}{100,000} \div \frac{9}{10} = \frac{999,999}{100,000} \times \frac{10}{9} = \frac{111,111}{10,000}$

The sum is: $\frac{111,111}{10,000}$

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Question

If the first term of a geometric sequence is 4, and the common ratio is $\frac{2}{3}$ , what is the fifth term?

Answer

Write the formula for the n-th term of the geometric sequence.

$a_n = a_1r^{n-1}$

Substitute the first term and common ratio.

The equation is: $a_n = 4(\frac{2}{3})^{n-1}$

To find the fifth term, substitute .

$a_5 = 4(\frac{2}{3})^{5-1} = 4(\frac{2}{3})^4 =4(\frac{2}{3})(\frac{2}{3})(\frac{2}{3})(\frac{2}{3})= 4(\frac{16}{81})$

The answer is: $\frac{64}{81}$

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Question

Given the sequence , what is the 7th term?

Answer

The formula for geometric sequences is defined by:

$ar^{n-1}$

The term represents the first term, while is the common ratio. The term represents the terms.

Substitute the known values.

$3(3)^{n-1}$

To determine the seventh term, simply substitute into the expression.

$3(3)^{7-1}$

The answer is:

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