### All SAT Math Resources

## Example Questions

### Example Question #1 : Arithmetic Sequences

-27, -24, -21, -18…

In the sequence above, each term after the first is 3 greater than the preceding term. Which of the following could not be a value in the sequence?

**Possible Answers:**

461

126

501

657

**Correct answer:**

461

All of the values in the sequence must be a multiple of 3. All answers are multiples of 3 except 461 so 461 cannot be part of the sequence.

### Example Question #1 : Arithmetic Sequences

*m, 3m, 5m, ...*

The first term in the above sequence is *m*, and each subsequent term is equal to *2m* + the previous term. If *m *is an integer, then which of the following could NOT be the sum of the first four terms in this sequence?

**Possible Answers:**

48

80

16

60

-32

**Correct answer:**

60

The fourth term of this sequence will be *5m* + *2m* = *7m*. If we add up the first four terms, we get *m + 3m + 5m + 7m = 4m + 12m = 16m*. Since *m* is an integer, the sum of the first four terms, *16m*, will have a factor of 16. Looking at the answer choices, 60 is the only answer where 16 is not a factor, so that is the correct choice.

### Example Question #1 : How To Find The Answer To An Arithmetic Sequence

The tenth term in a sequence is 40, and the twentieth term is 20. The difference between consequence terms in the sequence is constant. Find n such that the sum of the first n numbers in the sequence equals zero.

**Possible Answers:**

**Correct answer:**59

Let d represent the common difference between consecutive terms.

Let a_{n }denote the nth term in the sequence.

In order to get from the tenth term to the twentieth term in the sequence, we must add d ten times.

Thus a_{20 = }a_{10} + 10d

20 = 40 + 10d

d = -2

In order to get from the first term to the tenth term, we must add d nine times.

Thus a_{10 = }a_{1 + }9d

40 = a_{1 }+ 9(-2)

The first term of the sequence must be 58.

Our sequence looks like this: 58,56,54,52,50…

We are asked to find the nth term such that the sum of the first n numbers in the sequence equals 0.

58 + 56 + 54 + …. a_{n = }0

Eventually our sequence will reach zero, after which the terms will become the negative values of previous terms in the sequence.

58 + 56 + 54 + … 6 + 4 + 2 + 0 + -2 + -4 + -6 +….-54 + -56 + -58 = 0

The sum of the term that equals -2 and the term that equals 2 will be zero. The sum of the term that equals -4 and the term that equals 4 will also be zero, and so on.

So, once we add -58 to all of the previous numbers that have been added before, all of the positive terms will cancel, and we will have a sum of zero. Thus, we need to find what number -58 is in our sequence.

It is helpful to remember that a_{n }= a_{1 }+ d(n-1), because we must add d to a_{1 }exactly n-1 times in order to give us a_{n}. For example, a_{5} = a_{1} + 4d, because if we add d four times to the first term, we will get the fifth term. We can use this formula to find n.

-58 = a_{n} = a_{1} + d(n-1)

-58 = 58 + (-2)(n-1)

n = 59

### Example Question #4 : Arithmetic Sequences

The first term of a sequence is 1, and every term after the first term is –2 times the preceding term. How many of the first 50 terms of this sequence are less than 5?

**Possible Answers:**

27

8

54

64

16

**Correct answer:**

27

We can see how the sequence begins by writing out the first few terms:

1, –2, 4, –8, 16, –32, 64, –128.

Notice that every other term (of which there are exactly 50/2 = 25) is negative and therefore less than 25. Also notice that after the fourth term, every term is greater in absolute value than 5, so we just have to find the number of positive terms before the fourth term that are less than 5 and add that number to 25 (the number of negative terms in the first 50 terms).

Of the first four terms, there are only two that are less than 5 (i.e. 1 and 4), so we include these two numbers in our count: 25 negative numbers plus an additional 2 positive numbers are less than 5, so 27 of the first 50 terms of the sequence are less than 5.

### Example Question #1 : Arithmetic Sequences

**Possible Answers:**

**Correct answer:**

Look for cancellations to simplify. The sum of all consecutive integers from to is equal to . Therefore, we must go a little farther. , so the last number in the sequence in . That gives us negative integers, positive integers, and don't forget zero! .

### Example Question #1671 : Sat Mathematics

Brad can walk 3600 feet in 10 minutes. How many yards can he walk in ten seconds?

**Possible Answers:**

**Correct answer:**

If Brad can walk 3600 feet in 10 minutes, then he can walk 3600/10 = 360 feet per minute, and 360/60 = 6 feet per second.

There are 3 feet in a yard, so Brad can walk 6/3 = 2 yards per second, or 2 x 10 = 20 yards in 10 seconds.

### Example Question #1673 : Sat Mathematics

The first, third, fifth and seventh terms of an arithmetic sequence are , , and . Find the equation of the sequence where corresponds to the first term.

**Possible Answers:**

**Correct answer:**

The first important thing to note is that the way these answer choices are set up, any answer that does not have a at the end - that is, denoting first term of as specified - can be automatically eliminated. The second important thing is realizing that we are given terms that are not consecutive but are two apart, meaning we can use the usual common difference but need to halve it instead of taking it at face value (specifically, .)

### Example Question #1674 : Sat Mathematics

Find the unknown term in the sequence:

**Possible Answers:**

**Correct answer:**

The pattern in this sequence is , where represents the term's place in the sequence. It follows like so:

, our first term.

, our second term.

Then, our third term must be:

.

### Example Question #1 : How To Find The Answer To An Arithmetic Sequence

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

**Possible Answers:**

The sequence has no integers.

**Correct answer:**

The sequence has no integers.

Rewrite all three fractions in terms of their least common denominator, which is :

;

remains as is;

The sequence begins

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

Setting and ,

If this common difference is added a few more times, a pattern emerges:

...

All of the denominators end in 4 or 9, so none of them can be divisible by 20. Therefore, none of the terms will be integers.

### Example Question #1676 : Sat Mathematics

An arithmetic sequence begins as follows:

What is the first positive number in the sequence?

**Possible Answers:**

The nineteenth term

The twenty-second term

The twentieth term

The twenty-third term

The twenty-first term

**Correct answer:**

The twentieth term

Given the first two terms and , the common difference of an arithmetic sequence is equal to the difference:

Setting , :

The th term of an arithmetic sequence can be found by way of the formula

Since we are looking for the first positive number - equivalently, the first number greater than 0:

for some .

Setting and , and solving for :

Since must be a whole number, it follows that the least value of for which is ; therefore, the first positive term in the sequence is the twentieth term.