Sequences - SAT Math

Four consecutive integers could be represented as n, n+1, n+2, n+3

Therefore, by saying that they have a mean of 9.5, we mean to say:

(n + n+1 + n+2 + n+ 3)/4 = 9.5

(4n + 6)/4 = 9.5 → 4n + 6 = 38 → 4n = 32 → n = 8

Therefore, the largest value is n + 3, or 11.

Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108. Solving for x yields x=34. However, the question asks for the largest number, which is x+4 or 38. Please make sure to answer what the question asks for!

You could have also plugged in the answer choices. If you plugged in 38 as the largest number, then the previous even integer would be 36 and the next previous even integer 34. The sum of 34, 36, and 38 yields 108.

Consecutive odd integers can be represented as x, x+2, x+4, and x+6.

We know that the sum of these integers is 32. We can add the terms together and set it equal to 32:

x + (x+2) + (x+4) + (x+6) = 32

4x + 12 = 32

4x = 20

x = 5; x+2=7; x+4 = 9; x+6 = 11

Our integers are 5, 7, 9, and 11.

Let us call x the smallest integer. Because the next two numbers are consecutive even integers, we can call represent them as x + 2 and x + 4. We are told the sum of x, x+2, and x+4 is equal to 72.

x + (x + 2) + (x + 4) = 72

3x + 6 = 72

3x = 66

x = 22.

This means that the integers are 22, 24, and 26. The question asks us for the product of these numbers, which is 22(24)(26) = 13728.

The answer is 13728.

Let x be the smallest of the four integers. We are told that the integers are consecutive odd integers. Because odd integers are separated by two, each consecutive odd integer is two larger than the one before it. Thus, we can let x + 2 represent the second integer, x + 4 represent the third, and x + 6 represent the fourth. The sum of the four integers equals 96, so we can write the following equation:

x + (x + 2) + (x + 4) + (x + 6) = 96

Combine x terms.

4_x_ + 2 + 4 + 6 = 96

Combine constants on the left side.

4_x_ + 12 = 96

Subtract 12 from both sides.

4_x_ = 84

Divide both sides by 4.

x = 21

This means the smallest integer is 21. The other integers are therefore 23, 25, and 27.

The question asks us how many of the four integers are prime. A prime number is divisible only by itself and one. Among the four integers, only 23 is prime. The number 21 is divisible by 3 and 7; the number 25 is divisible by 5; and 27 is divisible by 3 and 9. Thus, 23 is the only number from the integers that is prime. There is only one prime integer.

The answer is 1.

Assume the three consecutive integers equal , , and . The sum of these three integers is 60. Thus,

There are 5 numbers in the sequnce.

How many numbers are left over if you divide 5 into 457?

There would be 2 numbers!

The second number in the sequence is 9,5,6,2,1

Since are consecutive integers, we know that at least 2 of them will be even. Since we have 2 that are going to be even, we know that when we divide the product by 2 we will still have an even number. Since 2 is the only prime that is even, we must have:

What we notice, however, is that for , we have the product is 0. For , we have the product is 24. We will then never have a product of 4, meaning that is never going to be a prime number.

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.