Sequences and Series - SSAT Upper Level Quantitative

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Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the next term of the arithmetic sequence:

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The common difference is . Add this to the last given term to find the next term.

Find the common difference for the arithmetic sequence:

$-$\frac{8}{3}$, -$\frac{2}{3}$, $\frac{4}{3}$, $\frac{10}{3}$...$

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Subtract the first term from the second term to find the common difference.

$-$\frac{2}{3}$-\left(-$\frac{8}{3}\right)=-$\frac{2}{3}$+\frac{8}{3}$=\frac{6}{3}$=2$

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

Tap to see back →

Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

Find the common difference for the arithmetic sequence:

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Subtract the first term from the second term to find the common difference.

An arithmetic sequence begins as follows:

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

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The common difference of the sequence is

so the th term of the sequence is

To find out the minimum value for which , set up this inequality:

$n > 302.7 \div 2.7 \approx 112.1$

The first negative term is the one hundred thirteenth term.

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?

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The common difference of the sequence is

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

so the th term of the sequence is

To find out the minimum value for which , 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}$$

The correct response is the forty-eighth term.

Three consecutive odd integers have sum 537. What is the product of the least and greatest of the three?

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Let the middle integer of the three be . The three integers are therefore

, and they can be found using the equation

$\left (x-2 \right )+ x +\left ( x+2 \right ) = 537$

The three integers are 177, 179, and 181, and the product of the least and greatest is

$177 \times 181 = 32,037$

Which of the following can be the sum of four consecutive positive integers?

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Let , , , and be the four consecutive integers. Then their sum would be

In other words, if 6 were to be subtracted from their sum, the difference would be a multiple of 4. Therefore, we subtract 6 from each of the choices and see if any of the resulting differences are multiples of 4.

$178 -6 = 172 ; 172 \div4 = 43$

$213 -6 = 207 ; 207 \div4 = 51 ; R ; 3$

$420 -6 = 414 ; 414 \div4 = 103 : R : 2$

$143 -6 = 137 ; 137 \div4 = 34: R: 1$

Since this only happens in the case of 178, this is the only number of the four that can be a sum of four consecutive integers: 43, 44, 45, 46.

Three consecutive integers have sum . What is their product?

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Let the middle integer of the three be . The three integers are therefore

, and they can be found using the equation

$\left (x-1 \right )+ x +\left ( x+1 \right ) = 427$

$x = 142$\frac{1}{3}$$

This contradicts the condition that the numbers are integers. Therefore, three integers satisfying the given conditions cannot exist.

Three consecutive even integers have sum 924. What is the product of the least and greatest of the three?

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Let the middle integer of the three be . The three integers are therefore

, and they can be found using the equation

$\left (x-2 \right )+ x +\left ( x+2 \right ) = 924$

The three even integers are therefore 306, 308, and 310, and the product of the least and greatest of these is

$306 \times 310= 94,860$

Four consecutive integers have sum 3,350. What is the product of the middle two?

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Call the least of the four integers . The four integers are therefore

,

and they can be found using the equation

$x + \left (x+1 \right )+\left (x+2 \right )+\left ( x+3 \right )= 3,350$

$x=3,344 \div 4 = 836$

The integers are 836, 837, 838, 839.

To get the correct response, multiply:

$837 \times 838 = 701,406$

Three consecutive integers have a sum of . What is their product?

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Let the middle integer of the three be . The three integers are therefore

, and they can be found using the equation

$\left (x-1 \right )+ x +\left ( x+1 \right ) = 312$

The integers are therefore 103, 104, 105. The correct response is their product, which is

$103 \times 104 \times 105= 1,124,760$