Arithmetic Sequences - SAT Math

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.

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.

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

Each number in the sequence in 7 more than the number preceding it.

The equation for the terms in an arithmetic sequence is an = a1 + d(n-1), where d is the difference.

The formula for the terms in this sequence is therefore an = 2 + 7(n-1).

Plug in 8 for n to find the 8th term:

a8 = 2 + 7(8-1) = 51

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

The answer is 15.6 .

Let _a_1 be the first term in the sequence. We can use the fact that a n+1 = (an)2 – 1 in order to find expressions for the second and third terms of the sequence in terms of _a_1.

_a_2 = (_a_1)2 – 1

_a_3 = (_a_2)2 – 1 = ((_a_1)2 – 1)2 – 1

We can use the fact that, in general, (a – b)2 = a_2 – 2_ab + _b_2 in order to simplify the expression for _a_3.

_a_3 = ((_a_1)2 – 1)2 – 1

= (_a_1)4 – 2(_a_1)2 + 1 – 1 = (_a_1)4 – 2(_a_1)2

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

_a_3 = (_a_1)2

We can substitute (_a_1)4 – 2(_a_1)2 for _a_3.

(_a_1)4 – 2(_a_1)2 = (_a_1)2

Subtract (_a_1)2 from both sides.

(_a_1)4 – 3(_a_1)2 = 0

Factor out (_a_1)2 from both terms.

(_a_1)2 ((_a_1)2 – 3) = 0

This means that either (_a_1)2 = 0, or (_a_1)2 – 3 = 0.

If (_a_1)2 = 0, then _a_1 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 (_a_1)2 – 3 = 0.

Add 3 to both sides.

(_a_1)2 = 3

Take the square root of both sides.

_a_1 = ±√3

However, since all the terms are positive, the only possible value for _a_1 is √3.

Now, that we know that _a_1 = √3, we can find _a_2, _a_3, and _a_4.

_a_2 = (_a_1)2 – 1 = (√3)2 – 1 = 3 – 1 = 2

_a_3 = (_a_2)2 – 1 = 22 – 1 = 4 – 1 = 3

_a_4 = (_a_3)2 – 1 = 32 – 1 = 9 – 1 = 8

The question ultimately asks for the product of the _a_2, _a_3, and _a_4, which would be equal to 2(3)(8), or 48.

The answer is 48.

From the given information, we know , which means each consecutive difference is 3.

The difference between each term can be found through subtraction. For example the difference between the first and the second term can be found as follows:

One can check and see that this is the case for the other given numbers in the sequence as well.

In order to find the seventh term, expand the sequence by adding 14 to the last given number (4th number) and all of the following numbers until the 7th number in the sequence is reached.

This gives the sequence:

As seen above the seventh number in the sequence is 87 and the correct answer.

The purpose of this question is to understand the patterns of sequences.

First, an equation for the term in the sequence must be determined ().

This is true because

will create ,

will create ,

will create ,

will create .

Then, the eqution must be applied to find the specified term. For the tenth term, the expression must be evaluated, yielding 103.

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:

Given the first two terms and , the common difference 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 four-digit whole number - equivalently, the first number greater than or equal to 1,000:

Setting and and solving for :

Therefore, the 77th term, or , is the first element in the sequence greater than 1,000. Substituting , , and in the rule and evaluating:

,

the correct choice.

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.

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.

Let d represent the common difference between consecutive terms.

Let an 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 a20 = a10 + 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 a10 = a1 + 9d

40 = a1 + 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 + …. an = 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 an = a1 + d(n-1), because we must add d to a1 exactly n-1 times in order to give us an. For example, a5 = a1 + 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 = an = a1 + d(n-1)

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

n = 59

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.

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! .

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.

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, .)

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:

.

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.