How to find the nth term of an arithmetic sequence

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SSAT Upper Level Quantitative › How to find the nth term of an arithmetic sequence

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An arithmetic sequence begins as follows:

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

The forty-first term

CORRECT

The forty-second term

The forty-third term

The forty-fourth term

The fortieth term

Explanation

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.

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?

The forty-eighth term

CORRECT

The forty-ninth term

The fiftieth term

The forty-seventh term

The fifty-first term

Explanation

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.

An arithmetic sequence begins as follows:

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

The fortieth term

CORRECT

The thirty-ninth term

The thirty-eighth term

The thirty-seventh term

The sequence has no positive terms.

Explanation

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.

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

CORRECT

Explanation

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

The common difference is

The seventieth term is

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

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.

The correct answer is not given among the other four responses.

CORRECT

$9\frac{4}{5}$

$10 \frac{1}{5}$

$10\frac{1}{10}$

$9\frac{9}{10}$

Explanation

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.

An arithmetic sequence begins as follows:

Give the thirty-second term of this sequence.

CORRECT

Explanation

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

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

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

An arithmetic sequence begins as follows:

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

The one hundred thirteenth term

CORRECT

The one hundred twelfth term

The one hundred eleventh term

The one hundred tenth term

The one hundred fourteenth term

Explanation

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.

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

CORRECT

$1\frac{1}{2}$

$-2\frac{1}{2}$

$-5\frac{1}{2}$

Explanation

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 $a_{1}$ :

$a_{11} = a_{1}+ (11-1) \frac{3}{2}$

$11= a_{1}+ 10 \cdot \frac{3}{2}$

$11= a_{1}+ 15$

$11- 15= a_{1}+ 15 - 15$

$a_{1} = -4$

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

CORRECT

Explanation

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

The hundredth term is

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

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.

1,936 square inches

CORRECT

484 square inches

2,304 square inches

576 square inches

784 square inches

Explanation

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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