How to find the nth term of an arithmetic sequence - 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

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

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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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