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SSAT Upper Level Math : Sequences and Series

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Example Questions

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Example Question #4 :n项的算术Sequence

An arithmetic sequence begins as follows:

5.7, 8.1,...

Which of the following is the first term greater than 100?

Possible Answers:

的fortieth term

的forty-first term

的forty-fourth term

的forty-second term

的forty-third term

Correct answer:

的forty-first term

Explanation:

的common difference of the sequence is

8.1 - 5.7 = 2.4

so theth term of the sequence is

$a_{n} =5.7+2.4 (n-1)$

To find out the minimum value for which $a_{n} > 100$ , set up this inequality:

5.7+2.4 (n-1) > 100

5.7+2.4 n-2.4 > 100

2.4 n +3.3 > 100

2.4 n > 96.7

$n > 96.7 \div 2.4 \approx 40.3$

的forty-first term is the correct response.

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Example Question #5 :n项的算术Sequence

An arithmetic sequence begins as follows:

300, 297.3,...

Which of the following terms is the first negative term in the sequence?

Possible Answers:

的one hundred eleventh term

的one hundred thirteenth term

的one hundred fourteenth term

的one hundred twelfth term

的one hundred tenth term

Correct answer:

的one hundred thirteenth term

Explanation:

的common difference of the sequence is

297.3 - 300 = -2.7

so theth term of the sequence is

$a_{n} = 300 -2.7 (n-1)$

To find out the minimum value for which $a_{n} < 0$ , set up this inequality:

300 -2.7 (n-1) < 0

300 -2.7 n+2.7 < 0

302.7-2.7 n < 0

302.7 < 2.7 n

2.7 n > 302.7

$n > 302.7 \div 2.7 \approx 112.1$

的first negative term is the one hundred thirteenth term.

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Example Question #6 :n项的算术Sequence

An arithmetic sequence begins as follows:

$155 \frac{5}{6}, 153 \frac{3}{4},...$

Which of the following terms is the first negative term in the sequence?

Possible Answers:

的seventy-fifth term

的seventy-eighth term

的seventy-sixth term

的seventy-seventh term

的seventy-fourth term

Correct answer:

的seventy-sixth term

Explanation:

的common difference of the sequence is

$153 \frac{3}{4}- 155 \frac{5}{6}= -2 \frac{1}{12}$

so theth term of the sequence is

$a_{n} = 155\frac{5}{6}-\left (2 \frac{1}{12} \right )(n-1)$

To find out the minimum value for which $a_{n} < 0$ , set up this inequality:

$155\frac{5}{6}-\left (2 \frac{1}{12} \right )(n-1) < 0$

$155\frac{5}{6}-\left (2 \frac{1}{12} \right ) n + 2\frac{1}{12} < 0$

$157\frac{11}{12}-\left (2 \frac{1}{12} \right ) n < 0$

$157\frac{11}{12} < \left (2 \frac{1}{12} \right ) n$

$\left (2 \frac{1}{12} \right ) n > 157\frac{11}{12}$

$n > 157\frac{11}{12} \div 2 \frac{1}{12} = 75 \frac{4}{5}$

的seventy-sixth term is the first negative term.

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Example Question #7 :n项的算术Sequence

An arithmetic sequence begins as follows:

$-78 \frac{2}{3}, -75 \frac{4}{5},...$

Which of the following terms is the first positive term in the sequence?

Possible Answers:

的twenty-ninth term

的twenty-seventh term

的twenty-eighth term

的sequence has no positive terms.

的thirtieth term

Correct answer:

的twenty-ninth term

Explanation:

的common difference of the sequence is

$-75 \frac{4}{5} - \left ( -78 \frac{2}{3} \right ) =2 \frac{13}{15}$ ,

so theth term of the sequence is

$a_{n} = - 78\frac{2}{3} + \left ( 2\frac{13}{15} \right ) (n-1)$

To find out the minimum value for which $a_{n} > 0$ , set up this inequality:

$- 78\frac{2}{3} + \left ( 2\frac{13}{15} \right ) (n-1) > 0$

$- 78\frac{2}{3} + \left ( 2\frac{13}{15} \right ) n - 2\frac{13}{15} > 0$

$\left ( 2\frac{13}{15} \right ) n - 81\frac{8}{15} > 0$

$\left ( 2\frac{13}{15} \right ) n > 81\frac{8}{15}$

$n > 81\frac{8}{15} \div 2\frac{13}{15} = 28 \frac{19}{43}$

的first positive term in the sequence is the twenty-ninth term.

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Example Question #8 :n项的算术Sequence

An arithmetic sequence begins as follows:

$1 \frac{1}{7}, 3 \frac{1}{4},...$

Which of the following is the first term greater than 100?

Possible Answers:

的forty-eighth term

的fiftieth term

的fifty-first term

的forty-seventh term

的forty-ninth term

Correct answer:

的forty-eighth term

Explanation:

的common difference of the sequence is

$3 \frac{1}{4} - 1 \frac{1}{7} = 2\frac{3}{28}$

so theth term of the sequence is

$a_{n} = 1\frac{1}{7} + \left ( 2\frac{3}{28} \right ) (n-1)$

To find out the minimum value for which $a_{n} > 100$ , set up this inequality:

$1\frac{1}{7} + \left ( 2\frac{3}{28} \right ) (n-1) > 100$

$1\frac{1}{7} + \left ( 2\frac{3}{28} \right ) n -2\frac{3}{28} > 100$

$\left ( 2\frac{3}{28} \right ) n - \frac{27}{28} > 100$

$\left ( 2\frac{3}{28} \right ) n > 100 \frac{27}{28}$

$n > 100 \frac{27}{28} \div 2\frac{3}{28} = 47 \frac{54}{59}$

的correct response is the forty-eighth term.

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Example Question #9 :n项的算术Sequence

的tenth and twelfth terms of an arithmetic sequence are 8.4 and 10.2. What is its first term?

Possible Answers:

1.8

-0.6

1.2

0.3

0.9

Correct answer:

0.3

Explanation:

的th term of an arithmetic sequence with initial term $a_{1}$ and common differenceis defined by the equation

$a_{n} = a_{1}+ (n-1)d$

Since the tenth and twelfth terms are two terms apart, the common difference can be found as follows:

$a_{10}+2d=a_{12}$

8.4+2d=10.2

8.4+2d- 8.4=10.2 - 8.4

2d=1.8

d = 0.9

Now, we can set n = 10, d = 0.9 in the sequence equation to find $a_{1}$ :

$a_{n} = a_{1}+ (n-1)d$

$a_{10} = a_{1}+ (10-1)0.9$

$8.4 = a_{1}+ 9 \cdot 0.9$

$8.4 = a_{1}+ 8.1$

$8.4 - 8.1 = a_{1}+ 8.1 - 8.1$

$a_{1} = 0.3$

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Example Question #10 :n项的算术Sequence

的eleventh and thirteenth terms of an arithmetic sequence are, respectively, 11 and 14. Give its first term.

Possible Answers:

$-5\frac{1}{2}$

$-2\frac{1}{2}$

$1\frac{1}{2}$

Correct answer:

Explanation:

的th term of an arithmetic sequence with initial term $a_{1}$ and common differenceis defined by the equation

$a_{n} = a_{1}+ (n-1)d$

Since the eleventh and thirteenth terms are two terms apart, the common difference can be found as follows:

$a_{11}+2d=a_{13}$

11+2d=14

11+2d -11=14 -11

2d = 3

$d = \frac{3}{2}$

Now, we can set $n = 11, d =\frac{3}{2}$ in the sequence equation to find $a_{1}$ :

$a_{11} = a_{1}+ (11-1) \frac{3}{2}$

$11= a_{1}+ 10 \cdot \frac{3}{2}$

$11= a_{1}+ 15$

$11- 15= a_{1}+ 15 - 15$

$a_{1} = -4$

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Example Question #11 :How To Find The Nth Term Of An Arithmetic Sequence

的lengths of the sides of ten squares form an arithmetic sequence. One side of the smallest square measures eight inches; one side of the second-smallest square measures one foot.

Give the area of the largest square.

Possible Answers:

784 square inches

484 square inches

576 square inches

1,936 square inches

2,304 square inches

Correct answer:

1,936 square inches

Explanation:

Let $a_{n}$ be the lengths of the sides of the squares in inches. $a_{1} = 8$ and $a_{2} = 12$ , so their common difference is

$d = a_{2} - a_{1} = 12 - 8 = 4$

的arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

的length of a side of the largest square - square 10 - can be found by substituting $a_{1} = 8, n= 10, d = 4$ :

$a_{10} =8+ (10-1) 4 = 8 + 9 \cdot 4 = 8 + 36 = 44$

的largest square has sides of length 44 inches, so its area is the square of this, or $A = 44^{2} = 1,936$ square inches.

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Example Question #12 :How To Find The Nth Term Of An Arithmetic Sequence

An arithmetic sequence begins as follows:

12, 19, 26, 33,...

Give the thirty-second term of this sequence.

Possible Answers:

236

229

243

222

250

Correct answer:

229

Explanation:

的th term of an arithmetic sequence with initial term $a_{1}$ and common differenceis defined by the equation

$a_{n} = a_{1}+ (n-1)d$

的initial term in the given sequence is

$a_{1} = 12$ ;

the common difference is

d = 19-12= 7 ;

We are seeking term n = 32 .

This term is

$a_{32} = 12+ (32-1) 7$

$a_{32} = 12+ 31 \cdot 7 = 12+ 217 = 229$

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Example Question #13 :How To Find The Nth Term Of An Arithmetic Sequence

An arithmetic sequence begins as follows:

$\frac{2}{5} , \frac{7}{10}, 1, 1 \frac{3}{10}, ...$

Give the thirty-third term of this sequence.

Possible Answers:

$9\frac{9}{10}$

的correct answer is not given among the other four responses.

$10\frac{1}{10}$

$10 \frac{1}{5}$

$9\frac{4}{5}$

Correct answer:

的correct answer is not given among the other four responses.

Explanation:

的th term of an arithmetic sequence with initial term $a_{1}$ and common differenceis defined by the equation

$a_{n} = a_{1}+ (n-1)d$ .

的initial term in the given sequence is

$a_{1} = \frac{2}{5}$ ;

the common difference is

$d = \frac{7}{10} - \frac{2}{5} = \frac{7}{10} - \frac{4}{10} = \frac{3}{10}$ .

We are seeking term n = 33 .

的refore,

$a_{33} = \frac{2}{5}+ (33-1) \frac{3}{10}$

$= \frac{2}{5}+ (32) \frac{3}{10}$

$= \frac{2}{5}+ \frac{96}{10}$

$= \frac{2}{5}+ \frac{48}{5}$

$= \frac{50}{5} = 10$ ,

which is not among the choices.

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SSAT Upper Level Math : Sequences and Series

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #4 :n项的算术Sequence

Example Question #5 :n项的算术Sequence

Example Question #6 :n项的算术Sequence

Example Question #7 :n项的算术Sequence

Example Question #8 :n项的算术Sequence

Example Question #9 :n项的算术Sequence

Example Question #10 :n项的算术Sequence

Example Question #11 :How To Find The Nth Term Of An Arithmetic Sequence

Example Question #12 :How To Find The Nth Term Of An Arithmetic Sequence

Example Question #13 :How To Find The Nth Term Of An Arithmetic Sequence

All SSAT Upper Level Math Resources

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