Sequences and Series - SSAT Upper Level Quantitative

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Question

The first two terms of an arithmetic sequence are 4 and 9, in that order. Give the one-hundredth term of that sequence.

Answer

The first term is $x_{0} = 4$ ; the common difference is

.

The hundredth term is

$x_{99} = x_{0} + 99 d = 4 + 99 \cdot5 = 4 + 495 = 499$ .

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Question

The first two terms of an arithmetic sequence are 1,000 and 997, in that order. What is the seventieth term?

Answer

The first term is $x_{0} = 1,000$ .

The common difference is

.

The seventieth term is

$x_{69} = x_{0}+ 69 \cdot d= 1,000 + 69 \cdot (-3)= 1,000 - 207 = 793$ .

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Question

An arithmetic sequence begins as follows:

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

Answer

The common difference of the sequence is

,

so the th term of the sequence is

$a_{n} = -128 + 3.3 (n-1)$

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

The first positive term is the fortieth term.

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Question

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?

Answer

The common difference of the sequence is

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

so the th 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:

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

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

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

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

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

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

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Question

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?

Answer

The common difference of the sequence is

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

so the th 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:

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

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

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

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

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

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

The seventy-sixth term is the first negative term.

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Question

An arithmetic sequence begins as follows:

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

Answer

The common difference of the sequence is

so the th 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:

$n > 302.7 \div 2.7 \approx 112.1$

The first negative term is the one hundred thirteenth term.

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Question

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?

Answer

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

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

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

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

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

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

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

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

The correct response is the forty-eighth term.

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Question

An arithmetic sequence begins as follows:

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

Answer

The common difference of the sequence is

so the th 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:

$n > 96.7 \div 2.4 \approx 40.3$

The forty-first term is the correct response.

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Question

An arithmetic sequence begins as follows:

Give the thirty-second term of this sequence.

Answer

The th term of an arithmetic sequence with initial term $a_{1}$ and common difference is defined by the equation

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

The initial term in the given sequence is

$a_{1} = 12$ ;

the common difference is

;

We are seeking term .

This term is

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

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

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Question

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.

Answer

The th term of an arithmetic sequence with initial term $a_{1}$ and common difference is defined by the equation

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

The 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 .

Therefore,

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

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

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

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

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

which is not among the choices.

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Question

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

Answer

The th term of an arithmetic sequence with initial term $a_{1}$ and common difference is 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:

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

Now, we can set 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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Question

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

Answer

The th term of an arithmetic sequence with initial term $a_{1}$ and common difference is 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:

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

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

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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Question

The 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.

Answer

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

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

The arithmetic sequence formula is

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

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

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

The 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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Question

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

Answer

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.

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Question

Three consecutive integers have sum . What is their product?

Answer

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.

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Question

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

Answer

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

$177 \times 181 = 32,037$

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Question

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

Answer

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

$306 \times 310= 94,860$

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Question

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

Answer

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:

$837 \times 838 = 701,406$

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Question

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

Answer

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

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

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Question

What is the value of is this sequence?

Answer

This is a geometric sequence since the pattern of the sequence is through multiplication.

You have to multiple each value by to get the next one.

The value before is so .

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