### All SAT Math Resources

## Example Questions

### Example Question #1 : Sequences

How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?

**Possible Answers:**

9

13

12

11

10

**Correct answer:**

12

The difference between each number in the series is 9. You can substract nine 11 times from 100 to get 1: 100 – 9x11 = 1. Counting 100, there are 12 positive numbers in the series.

### Example Question #11 : Sequences

In a sequence of numbers, each term is times larger than the one before it. If the 3^{rd} term of the sequence is 12, and the 6^{th} term is 96, what is the sum of all of the terms less than 250?

**Possible Answers:**

192

384

381

378

372

**Correct answer:**

381

Let's call the first term in the sequence a_{1} and the nth term a_{n}.

We are told that each term is r times larger than the one before it. Thus, we can find the next term in the sequence by multiplying by r.

a_{1} = a_{1}_{}

a_{2} = r(a_{1})

a_{3} = r(a_{2}) = r(r(a_{1})) = r^{2}(a_{1})

a_{4} = r(a_{3}) = r(r^{2}(a_{1})) = r^{3}(a_{1})

a_{n} = r^{(n–1)}a_{1}

We can use this information to find r.

The problem gives us the value of the third and the sixth terms.

a_{3} = r^{2}(a_{1}) = 12

a_{6} = r^{5}(a_{1}) = 96

Let's solve for a_{1} in terms of r and a_{3}.

a_{1} = 12/(r^{2})

Let's then solve for a_{1} in terms of r and a_{6}.

a_{1} = 96/(r^{5})

Now, we can set both values equal and solve for r.

12/(r^{2}) = 96/(r^{5})

Multiply both sides by r^{5} to get rid of the fraction.

12r^{5}/r^{2} = 96

Apply the property of exponents which states that a^{b}/a^{c} = a^{b–c}.

12r^{3} = 96

Divide by 12 on both sides.

r^{3} = 8

Take the cube root of both sides.

r = 2

This means that each term is two times larger than the one before it, or that each term is one half as large as the one after it.

a_{2} must equal a_{3} divided by 2, which equals ^{12}/_{2} = 6.

a_{1} must equal a_{2} divided by 2, which equals ^{6}/_{2} = 3.

Here are the first eight terms of the sequence:

3, 6, 12, 24, 48, 96, 192, 384

The question asks us to find the sum of all the terms less than 250. Only the first seven terms are less than 250. Thus the sum is equal to the following:

sum = 3 + 6 + 12 + 24 + 48 + 96 + 192 = 381

### Example Question #12 : Sequences

Find the term of the sequence above.

**Possible Answers:**

**Correct answer:**

The sequence is geometric with a common ratio of .

The formula for finding the term of the sequence is

.

So

.

### Example Question #13 : Sequences

Which of the following are not natural numbers?

I. 1

II. 0

III. 349010

IV. -2

V. 1/4

**Possible Answers:**

I, IV, V

IV, V

II, III, IV, V

I, V

II, IV, V

**Correct answer:**

II, IV, V

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

### Example Question #14 : Sequences

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

**Possible Answers:**

The sequence has no integers.

**Correct answer:**

Subtract the first term from the second term to get the common difference :

Setting and ,

If is in the sequence, then there is an integer such that

, or

Solving for ,

Therefore, we seek the least positive integer value of such that is itself an integer. By trial and error, we see:

:

,

which is an integer.

Therefore, 2 is the first integer value in the sequence.

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