SAT Math - Integers | Practice Hub

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:

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

$= 12.8 + 15 \cdot 4.4$

What is the least common multiple of ?

Explanation

Least common multiple is the smallest number that is divisible by two or more factors. Since are prime numbers and can't be broken down to smaller factors, we just multiply them to get as our answer.

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.

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.

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.

Shannon decided to go to nearby café for lunch. She can have a sandwich made on either wheat or white bread. The café offers cheddar, Swiss, and American for cheese choices. For meat, Shannon can choose ham, turkey, bologna, roast beef, or salami. How many cheese and meat sandwich options does Shannon have to choose from?

Explanation

2 bread choices * 3 cheese choices * 5 meat choices = 30 sandwich choices

An ice cream parlor serves 36 ice cream flavors. You can order any flavor in a small, medium or large and can choose between a waffle cone and a cup. How many possible combinations could you possibly order?

108

144

172

216

Explanation

36 possible flavors * 3 possible sizes * 2 possible cones = 216 possible combinations.

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.

Integers

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