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Algebra II : Algebra II

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Example Questions

Example Question #6 : Mathematical Relationships And Basic Graphs

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

[the first term is 7,000]

Possible Answers:

0.000481

1.119

14,551, 915.228

0.000192

0.00000000184

Correct answer:

0.000481

Explanation:

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 7,000 , 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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Example Question #9 : Mathematical Relationships And Basic Graphs

Find the 21st term of the sequence 90, 108, 129.6, 155.52, ...

[90 is 1st, so n=1]

Possible Answers:

2.348

55.206

3,450.384

1.956

4,140.461

Correct answer:

3,450.384

Explanation:

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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Example Question #11 : Summations And Sequences

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

Possible Answers:

-11,738.206

-3.268

3.268

9,603.987

11,738.206

Correct answer:

-11,738.206

Explanation:

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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Example Question #2671 : Algebra Ii

Find the common ratio for this geometric series:

1,7,49,343,2401

Possible Answers:

343

$\frac{1}{7}$

Correct answer:

Explanation:

Find the common ratio for this geometric series:

1,7,49,343,2401

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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Example Question #12 : Summations And Sequences

Find the next term in this geometric series:

1,7,49,343,2401

Possible Answers:

16814

16807

343

Correct answer:

16807

Explanation:

Find the next term in this geometric series:

1,7,49,343,2401

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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Example Question #14 : Summations And Sequences

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

Possible Answers:

$\frac{4}{7}$

$\frac{1}{5}$

$\frac{6}{11}$

$\frac{1}{2}$

$\frac{3}{8}$

Correct answer:

$\frac{4}{7}$

Explanation:

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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Example Question #13 : Summations And Sequences

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

Possible Answers:

$\frac{64}{3}$

$\frac{20}{3}$

Correct answer:

$\frac{64}{3}$

Explanation:

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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Example Question #14 : Summations And Sequences

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

Possible Answers:

$3.495\times 10^{-5}$

$8.338\times 10^{-4}$

$4.127\times 10^{-7}$

$8.738\times 10^{-5}$

$1.776\times 10^{-9}$

Correct answer:

$8.738\times 10^{-5}$

Explanation:

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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Example Question #12 : Geometric Sequences

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

Possible Answers:

$\frac{111,111}{10,000}$

$\frac{111,111}{20,000}$

$\frac{10,000}{999}$

$\frac{100}{9}$

$\frac{11,111}{9,999}$

Correct answer:

$\frac{111,111}{10,000}$

Explanation:

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: n=6

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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Example Question #15 : Summations And Sequences

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

Possible Answers:

$\frac{64}{81}$

$\frac{325}{27}$

$\textup{The answer is not given.}$

$\frac{964}{81}$

$\frac{844}{81}$

Correct answer:

$\frac{64}{81}$

Explanation:

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 n=5 .

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

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

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #6 : Mathematical Relationships And Basic Graphs

Example Question #9 : Mathematical Relationships And Basic Graphs

Example Question #11 : Summations And Sequences

Example Question #2671 : Algebra Ii

Example Question #12 : Summations And Sequences

Example Question #14 : Summations And Sequences

Example Question #13 : Summations And Sequences

Example Question #14 : Summations And Sequences

Example Question #12 : Geometric Sequences

Example Question #15 : Summations And Sequences

All Algebra II Resources

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