# SAT Math : Sequences

## Example Questions

### Example Question #1 : Other Arithmetic Sequences

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.

Explanation:

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

Find the unknown term in the sequence:

Explanation:

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 #4 : How To Find The Answer To An Arithmetic Sequence

An arithmetic sequence begins as follows:

Give the first integer in the sequence.

The sequence has no integers.

The sequence has no integers.

Explanation:

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 #5 : How To Find The Answer To An Arithmetic Sequence

An arithmetic sequence begins as follows:

What is the first positive number in the sequence?

The twenty-first term

The nineteenth term

The twenty-third term

The twentieth term

The twenty-second term

The twentieth term

Explanation:

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.

### Example Question #791 : Arithmetic

An arithmetic sequence begins as follows:

Give the thirteenth term of the sequence as a fraction in lowest terms.

Explanation:

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

Setting  and ,

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

Setting , and  in the formula:

As a fraction, this is

### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence

2, 8, 14, 20

The first term in the sequence is 2, and each following term is determined by adding 6. What is the value of the 50th term?

302

300

296

320

296

Explanation:

We start by multiplying 6 times 46, since the first 4 terms are already listed. We then add the product, 276, to the last listed term, 20. This gives us our answer of 296.

### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence

Which of the following could not be a term in the sequence 5, 10, 15, 20...?

3751

10005

35

2500

3751

Explanation:

All answers in the sequence must end in a 5 or a 0.

### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence

In an arithmetic sequence, each term is two greater than the one that precedes it. If the sum of the first five terms of the sequence is equal to the difference between the first and fifth terms, what is the tenth term of the sequence?

–0.6

10

2.4

15.6

0.6

15.6

Explanation:

Let a1 represent the first term of the sequence and an represent the nth term.

We are told that each term is two greater than the term that precedes it. Thus, we can say that:

a2 = a1 + 2

a3 = a1 + 2 + 2 = a1 + 2(2)

a4 = a1 + 3(2)

a5 = a1 + 4(2)

an = a1 + (n-1)(2)

The problem tells us that the sum of the first five terms is equal to the difference between the fifth and first terms. Let's write an expression for the sum of the first five terms.

sum = a1 + (a1 + 2) + (a1 + 2(2)) + (a1 + 3(2)) + (a1 + 4(2))

= 5a1 + 2 + 4 + 6 + 8

= 5a1 + 20

Next, we want to write an expression for the difference between the fifth and first terms.

a5 - a1 = a1 + 4(2) – a1 = 8

Now, we set the two expressions equal and solve for a1.

5a1 + 20 = 8

Subtract 20 from both sides.

5a1 = –12

a1 = –2.4.

The question ultimately asks us for the tenth term of the sequence. Now, that we  have the first term, we can find the tenth term.

a10 = a1 + (10 – 1)(2)

a10 = –2.4 + 9(2)

= 15.6

### Example Question #1 : Nth Term Of An Arithmetic Sequence

In a certain sequence, an+1 = (an)2 – 1, where an represents the nth term in the sequence. If the third term is equal to the square of the first term, and all of the terms are positive, then what is the value of (a2)(a3)(a4)?

6

48

24

72

63

48

Explanation:

Let a1 be the first term in the sequence. We can use the fact that an+1 = (an)2 – 1 in order to find expressions for the second and third terms of the sequence in terms of a1.

a2 = (a1)2 – 1

a3 = (a2)2 – 1 = ((a1)2 – 1)2 – 1

We can use the fact that, in general, (a – b)2 = a2 – 2abb2 in order to simplify the expression for a3.

a= ((a1)2 – 1)2 – 1

= (a1)4 – 2(a1)2 + 1 – 1 = (a1)4 – 2(a1)2

We are told that the third term is equal to the square of the first term.

a3 = (a1)2

We can substitute (a1)4 – 2(a1)for a3.

(a1)4 – 2(a1)= (a1)2

Subtract (a1)2 from both sides.

(a1)4 – 3(a1)= 0

Factor out (a1)from both terms.

(a1)2 ((a1)2 – 3) = 0

This means that either (a1)= 0, or (a1)2 – 3 = 0.

If (a1)= 0, then a1 must be 0. However, we are told that all the terms of the sequence are positive. Therefore, the first term can't be 0.

Next, let's solve (a1)2 – 3 = 0.

(a1)= 3

Take the square root of both sides.

a1 = ±√3

However, since all the terms are positive, the only possible value for a1 is √3.

Now, that we know that a1 = √3, we can find a2, a3, and a4.

a2 = (a1)2 – 1 = (√3)2 – 1 = 3 – 1 = 2

a3 = (a2)2 – 1 = 2– 1 = 4 – 1 = 3

a4 = (a3)2 – 1 = 32 – 1 = 9 – 1 = 8

The question ultimately asks for the product of the a2, a3, and a4, which would be equal to 2(3)(8), or 48.

### Example Question #1 : Nth Term Of An Arithmetic Sequence

In the given sequence, the first term is 3 and each term after is one less than three times the previous term.

What is the sixth term in the sequence?

Explanation:

The fourth term is: 3(23) – 1 = 69 – 1 = 68.

The fifth term is: 3(68) – 1 = 204 – 1 = 203.

The sixth term is: 3(203) – 1 = 609 – 1 = 608.