Sequences

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SAT Math › Sequences

Questions 1 - 10

An arithmetic sequence begins as follows:

Give the sixteenth term of this sequence.

None of the other responses give the correct answer.

Explanation

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{1} = 12 .8$ and $a_{2} = 17.2$

The th term of an arithmetic sequence $a_{n}$ can be found by way of the formula

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

Setting , $a_{1} = 12 .8$ , and in the formula:

$a_{16} = 12.8 + (16-1) 4.4$

$= 12.8 + 15 \cdot 4.4$

An arithmetic sequence begins as follows:

What is the first positive number in the sequence?

The twentieth term

The twenty-first term

The twenty-second term

The twenty-third term

The nineteenth term

Explanation

Given the first two terms $a_{1}$ and $a_{2}$ , the common difference of an arithmetic sequence is equal to the difference:

$d = a_{2} - a_{1}$

Setting $a_{1} = -129$ , $a_{2} =-122$ :

The th term of an arithmetic sequence $a_{n}$ can be found by way of the formula

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

Since we are looking for the first positive number - equivalently, the first number greater than 0:

$a_{n} = a_{1}+ (n-1) d > 0$ for some .

Setting $a_{1} = -129$ and , and solving for :

$\frac{7n}{7} > \frac{136}{7}$

$n > 19\frac{3}{7}$

Since must be a whole number, it follows that the least value of for which $a_{n}> 0$ is ; therefore, the first positive term in the sequence is the twentieth term.

An arithmetic sequence begins as follows:

$\frac{1}{5} , \frac{9}{20} , \frac{7}{10}, ...$

Give the first integer in the sequence.

The sequence has no integers.

Explanation

Rewrite all three fractions in terms of their least common denominator, which is :

$\frac{1}{5} =\frac{1 \cdot 4}{5 \cdot 4} = \frac{4}{20}$ ;

$\frac{9}{20}$ remains as is;

$\frac{7}{10} = \frac{7 \cdot 2}{10 \cdot 2} = \frac{14}{20}$

The sequence begins

$\frac{4}{20},\frac{9}{20}, \frac{14}{20}...$

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = \frac{9}{20}$ and $a_{1} = \frac{4}{20}$ ,

$d = \frac{9}{20} -\frac{4}{20} = \frac{5}{20}$

If this common difference is added a few more times, a pattern emerges:

$\frac{14}{20} + \frac{5}{20} = \frac{19}{20}$

$\frac{19}{20} + \frac{5}{20} = \frac{24}{20}$ ...

$\frac{4}{20},\frac{9}{20}, \frac{14}{20} ,\frac{19}{20}, \frac{24}{20} ...$

All of the denominators end in 4 or 9, so none of them can be divisible by 20. Therefore, none of the terms will be integers.

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

The sequence has no integers.

Explanation

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = 0.05$ and $a_{1} = 0. 02$ ,

If is in the sequence, then there is an integer such that

$a= a_{1}+ N d$ , or

$a= 0.02 + N \cdot 0.03$

Solving for ,

$a- 0.02 = 0.02 + N \cdot 0.03 - 0.02$

$a- 0.02 = N \cdot 0.03$

$\frac{ a- 0.02 }{0.03} = \frac{N \cdot 0.03 }{0.03}$

$N = \frac{ a- 0.02 }{0.03}$

$N = \frac{100( a- 0.02) }{100(0.03)}$

$N = \frac{100 a-2 }{3}$

Therefore, we seek the least positive integer value of such that $N = \frac{100 a-2 }{3}$ is itself an integer. By trial and error, we see:

$N = \frac{100 \cdot 1 -2 }{3} = \frac{100 -2 }{3} = \frac{98}{3} = 32\frac{2}{3}$

$N = \frac{100 \cdot 2 -2 }{3} = \frac{200 -2 }{3} = \frac{198}{3} = 66$ ,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

The sequence has no integers.

Explanation

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = 0.05$ and $a_{1} = 0. 02$ ,

If is in the sequence, then there is an integer such that

$a= a_{1}+ N d$ , or

$a= 0.02 + N \cdot 0.03$

Solving for ,

$a- 0.02 = 0.02 + N \cdot 0.03 - 0.02$

$a- 0.02 = N \cdot 0.03$

$\frac{ a- 0.02 }{0.03} = \frac{N \cdot 0.03 }{0.03}$

$N = \frac{ a- 0.02 }{0.03}$

$N = \frac{100( a- 0.02) }{100(0.03)}$

$N = \frac{100 a-2 }{3}$

Therefore, we seek the least positive integer value of such that $N = \frac{100 a-2 }{3}$ is itself an integer. By trial and error, we see:

$N = \frac{100 \cdot 1 -2 }{3} = \frac{100 -2 }{3} = \frac{98}{3} = 32\frac{2}{3}$

$N = \frac{100 \cdot 2 -2 }{3} = \frac{200 -2 }{3} = \frac{198}{3} = 66$ ,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

An arithmetic sequence begins as follows:

Give the sixteenth term of this sequence.

None of the other responses give the correct answer.

Explanation

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{1} = 12 .8$ and $a_{2} = 17.2$

The th term of an arithmetic sequence $a_{n}$ can be found by way of the formula

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

Setting , $a_{1} = 12 .8$ , and in the formula:

$a_{16} = 12.8 + (16-1) 4.4$

$= 12.8 + 15 \cdot 4.4$

An arithmetic sequence begins as follows:

$\frac{1}{5} , \frac{9}{20} , \frac{7}{10}, ...$

Give the first integer in the sequence.

The sequence has no integers.

Explanation

Rewrite all three fractions in terms of their least common denominator, which is :

$\frac{1}{5} =\frac{1 \cdot 4}{5 \cdot 4} = \frac{4}{20}$ ;

$\frac{9}{20}$ remains as is;

$\frac{7}{10} = \frac{7 \cdot 2}{10 \cdot 2} = \frac{14}{20}$

The sequence begins

$\frac{4}{20},\frac{9}{20}, \frac{14}{20}...$

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{2} = \frac{9}{20}$ and $a_{1} = \frac{4}{20}$ ,

$d = \frac{9}{20} -\frac{4}{20} = \frac{5}{20}$

If this common difference is added a few more times, a pattern emerges:

$\frac{14}{20} + \frac{5}{20} = \frac{19}{20}$

$\frac{19}{20} + \frac{5}{20} = \frac{24}{20}$ ...

$\frac{4}{20},\frac{9}{20}, \frac{14}{20} ,\frac{19}{20}, \frac{24}{20} ...$

All of the denominators end in 4 or 9, so none of them can be divisible by 20. Therefore, none of the terms will be integers.

An arithmetic sequence begins as follows:

What is the first positive number in the sequence?

The twentieth term

The twenty-first term

The twenty-second term

The twenty-third term

The nineteenth term

Explanation

Given the first two terms $a_{1}$ and $a_{2}$ , the common difference of an arithmetic sequence is equal to the difference:

$d = a_{2} - a_{1}$

Setting $a_{1} = -129$ , $a_{2} =-122$ :

The th term of an arithmetic sequence $a_{n}$ can be found by way of the formula

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

Since we are looking for the first positive number - equivalently, the first number greater than 0:

$a_{n} = a_{1}+ (n-1) d > 0$ for some .

Setting $a_{1} = -129$ and , and solving for :

$\frac{7n}{7} > \frac{136}{7}$

$n > 19\frac{3}{7}$

Since must be a whole number, it follows that the least value of for which $a_{n}> 0$ is ; therefore, the first positive term in the sequence is the twentieth term.

An arithmetic sequence begins as follows:

Give the sixteenth term of this sequence.

None of the other responses give the correct answer.

Explanation

Subtract the first term $a_{1}$ from the second term $a_{2}$ to get the common difference :

$d = a_{2} - a_{1}$

Setting $a_{1} = 12 .8$ and $a_{2} = 17.2$

The th term of an arithmetic sequence $a_{n}$ can be found by way of the formula

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

Setting , $a_{1} = 12 .8$ , and in the formula:

$a_{16} = 12.8 + (16-1) 4.4$

$= 12.8 + 15 \cdot 4.4$

An arithmetic sequence begins as follows: 14, 27, 40...

What is the first four-digit integer in the sequence?

Explanation

Given the first two terms $a_{1}$ and $a_{2}$ , the common difference is equal to the difference:

$d = a_{2} - a_{1}$

Setting $a_{1} = 14$ , $a_{2} =27$ :

The th term of an arithmetic sequence $a_{n}$ can be found by way of the formula

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

Since we are looking for the first four-digit whole number - equivalently, the first number greater than or equal to 1,000:

$a_{1}+ (n-1) d \ge 1,000$

Setting $a_{1} = 14$ and and solving for :

$14 + (n-1) 13 \ge 1,000$

$14 + 13n-13 \ge 1,000$

$13n+1 \ge 1,000$

$13n+1 - 1 \ge 1,000 - 1$

$13 n \ge 999$

$13 n \div 13 \ge 999 \div 13$

$n \ge 76\frac{11}{13}$

Therefore, the 77th term, or $a_{77}$ , is the first element in the sequence greater than 1,000. Substituting , $a_{1} = 14$ , and in the rule and evaluating:

$a_{77}=a_{1}+ (n-1) d$

$= 14+76 \cdot 13$

the correct choice.

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