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SAT II Math II : Sequences

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #1 : Sequences

Evaluate:

$\sum_{i = 1}^{\infty} \left (\frac{4}{3} \right )^{i}$

Possible Answers:

$\frac{7}{4}$

The series diverges

$\frac{5}{3}$

$\frac{5}{4}$

Correct answer:

The series diverges

Explanation:

An infinite series $\sum_{i = 1}^{\infty}a^{i}$ converges to a sum if and only if |a| < 1 . However, in the series $\sum_{i = 1}^{\infty} \left (\frac{4}{3} \right )^{i}$ , this is not the case, as $\left | \frac{4}{3} \right |= \frac{4}{3} > 1$ . This series diverges.

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Example Question #2 : Sequences

Give the next term in this sequence:

$0,0, \frac{1}{2}, \frac{2}{3}, \frac{4}{5}, \frac{7}{8}, \frac{12}{13}, \frac{20}{21},$ _______________

Possible Answers:

$\frac{31}{32}$

$\frac{25}{26}$

$\frac{33}{34}$

$\frac{34}{35}$

Correct answer:

$\frac{33}{34}$

Explanation:

The key to finding the next term lies in the denominators of the third term onwards. They are terms of the Fibonacci sequence, which begin with the terms 1 and 1 and whose subsequent terms are each formed by adding the previous two.

The th term of the sequence is the number $\frac{F_{n}-1}{F_{n}}$ , where $F_{n}$ is the th number in the Fibonacci sequence (since the first two Fibonacci numbers are both 1, the first two terms being 0 fits this pattern). The Fibonacci number following 13 and 21 is their sum, 34, so the next number in the sequence is

$\frac{34-1}{34}= \frac{33}{34}$ .

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Example Question #1 : Sequences

Give the next term in this sequence:

12, 13, 17, 26, 42, 67, __________

Possible Answers:

116

103

100

Correct answer:

103

Explanation:

$12\textbf{ + 1} = 13$

$13\textbf{ + 4}= 17$

$17\textbf{ + 9}= 26$

$26\textbf{ + 16} = 42$

$42\textbf{ + 25}=67$

Each term is derived from the next by adding a perfect square integer; the increment increases from one square to the next higher one each time. To maintain the pattern, add the next perfect square, 36:

$67\textbf{ + 36 } =103$

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Example Question #3 : Sequences

Give the next term in this sequence:

1, 1,3, 5, 11, 21, 43, 85, 171, _____________

Possible Answers:

343

339

340

341

342

Correct answer:

341

Explanation:

Each term is derived from the previous term by doubling the latter and alternately adding and subtracting 1, as follows:

$1 \textbf{ x 2 - 1 } = 1$

$1 \textbf{ x 2 + 1 } = 3$

$3 \textbf{ x 2 - 1 } = 5$

$5 \textbf{ x 2 + 1 } = 11$

$11 \textbf{ x 2 - 1 } =21$

$21 \textbf{ x 2 + 1 } = 43$

$43\textbf{ x 2 - 1 } = 85$

$85 \textbf{ x 2 + 1 } = 171$

The next term is derived as follows:

$171 \textbf{ x 2 - 1 } = 341$

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Example Question #111 : Mathematical Relationships

Give the next term in this sequence:

$1, \sqrt{2}, \sqrt{6}, 2\sqrt{6}, 2 \sqrt{30}, 12 \sqrt{5},$ _____________

Possible Answers:

$84\sqrt{5}$

$72\sqrt{5}$

The correct answer is not among the other responses.

$12 \sqrt{35}$

$12\sqrt{30}$

Correct answer:

$12 \sqrt{35}$

Explanation:

The pattern becomes more clear if each term is rewritten as a single radical expression:

$1 = \sqrt {1} = \sqrt{1!}$

$\sqrt{2} = \sqrt{2!}$

$\sqrt{6} = \sqrt{3!}$

$2\sqrt{6} =\sqrt{4} \cdot \sqrt{6} = \sqrt {24}=\sqrt{4!}$

$2 \sqrt{30}= \sqrt{4} \cdot \sqrt{30}= \sqrt{120} = \sqrt {5!}$

$12\sqrt{5} = \sqrt{144} \cdot \sqrt{5} = \sqrt{720} = \sqrt{6!}$

The th term is $\sqrt{n!}$ . The next (seventh) term is therefore

$\sqrt{7!}= \sqrt{5,040}= \sqrt{144} \times \sqrt{35} = 12 \sqrt{35}$

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Example Question #111 : Mathematical Relationships

A geometric sequence begins as follows:

6 , i ,...

Give the next term of the sequence.

Possible Answers:

6+ i

$\frac{1}{6}$

-6 + 2i

$-\frac{1}{6}$

6 - i

Correct answer:

$-\frac{1}{6}$

Explanation:

The common ratio of a geometric sequence is the quotient of the second term and the first:

$r = \frac{a_{2}}{a_{1}}= \frac{i}{6} = \frac{1}{6} i$

Multiply the second term by the common ratio to obtain the third term:

$a_{3} = r a_{2} = \frac{1}{6} i \cdot i = \frac{1}{6} \cdot i^{2}= \frac{1}{6} (-1)= -\frac{1}{6}$

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Example Question #141 : Sat Subject Test In Math Ii

A geometric sequence begins as follows:

$0.8 , \frac{6}{7} ...$

Give the next term of the sequence.

Possible Answers:

$\frac{32}{35}$

$\frac{8}{9}$

$\frac{45}{49}$

$1 \frac{3}{25}$

Correct answer:

$\frac{45}{49}$

Explanation:

Rewrite the first term as a fraction:

$0.8 = \frac{8}{10} = \frac{4}{5}$

The common ratio of a geometric sequence can be found by dividing the second term by the first, so

$r = \frac{6}{7} \div \frac{4}{5} = \frac{6}{7} \cdot \frac{5}{4} = \frac{15}{14 }$

The third term is equal to the second term multiplied by this common ratio:

$\frac{6}{7} \cdot \frac{15}{14 } = \frac{45}{49}$ .

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Example Question #1 : Sequences

A geometric sequence begins as follows:

1+ i, 2, ...

Give the next term of the sequence.

Possible Answers:

3- i

4 - i

2 - 2i

2+ 2i

4+i

Correct answer:

2 - 2i

Explanation:

The common ratio of a geometric sequence is the quotient of the second term and the first:

$r = \frac{a_{2}}{a_{1}}= \frac{2}{1+i }$

Multiply this common ratio by the second term to get the third term:

$a_{3} =r \cdot a_{2} = \frac{2}{1+i } \cdot 2 = \frac{4}{1+i }$

This can be expressed in standard form by rationalizing the denominator; do this by multiplying numerator and denominator by the complex conjugate of the denominator, which is 1 - i :

$\frac{4}{1+i }$

$= \frac{4(1-i)}{(1+i) (1-i) }$

$= \frac{4(1-i)}{1^{2}-i^{2}}$

$= \frac{4(1-i)}{1-(-1)}$

$= \frac{4(1-i)}{2}$

= 2(1-i)

= 2- 2i

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Example Question #1 : Sequences

A geometric sequence begins as follows:

$\sqrt{40} , - \sqrt{120},...$

Express the next term of the sequence in simplest radical form.

Possible Answers:

$6 \sqrt{10}$

$10 \sqrt{2}$

$-2 \sqrt{70}$

Correct answer:

$6 \sqrt{10}$

Explanation:

The common ratio of a geometric sequence is the quotient of the second term and the first. Using the Quotient of Radicals property, we can obtain:

$r = \frac{a_{2}}{a_{1}}= \frac{ -\sqrt{120}}{ \sqrt{40}} = - \sqrt{\frac{120}{40}} = -\sqrt{3}$

Multiply the second term by the common ratio, then simplify using the Product Of Radicals Rule, to obtain the third term:

$a_{3} = r a_{2}$

$=-\sqrt{ 3 }\cdot( - \sqrt{120} )$

$= \sqrt{ 3 }\cdot\sqrt{120}$

$= \sqrt{360 }$

$= \sqrt{36} \cdot \sqrt{10 }$

$= 6 \sqrt{10}$

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Example Question #1 : Sequences

The first and second terms of a geometric sequence are $\sqrt[3]{16}$ and $\sqrt[3]{48}$ , respectively. In simplest form, which of the following is its third term?

Possible Answers:

$4 \sqrt[3]{5}$

$2 \sqrt[3]{10}$

$2 \sqrt[3]{18}$

Correct answer:

$2 \sqrt[3]{18}$

Explanation:

The common ratio of a geometric sequence can be determined by dividing the second term by the first. Doing this and using the Quotient of Radicals Rule to simplfy:

$r = \frac{ a_{2}}{ a_{1}} = \frac{ \sqrt[3]{48}}{ \sqrt[3]{16}} = \sqrt[3]{ \frac{48}{16}} = \sqrt[3]{3}$

Multiply this by the second term to get the third term, simplifying using the Product of Radicals Rule

$a_{3} = r a_{2} = \sqrt[3]{3} \cdot \sqrt[3]{48}= \sqrt[3]{144} = \sqrt[3]{8} \cdot \sqrt[3]{18} = 2 \sqrt[3]{18}$

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SAT II Math II : Sequences

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : Sequences

Example Question #2 : Sequences

Example Question #1 : Sequences

Example Question #3 : Sequences

Example Question #111 : Mathematical Relationships

Example Question #111 : Mathematical Relationships

Example Question #141 : Sat Subject Test In Math Ii

Example Question #1 : Sequences

Example Question #1 : Sequences

Example Question #1 : Sequences

All SAT II Math II Resources

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