Sequences - PSAT Math

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.

The pattern of the sequence is (x+1) * 2.

We have the first 5 terms, so we need terms 6 and 7:

(78+1) * 2 = 158

(158+1) * 2 = 318

3 + 8 + 18 +38 + 78 + 158 + 318 = 621

The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.

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.

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.

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.

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.

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.

Shortcut: If you add up al the integers from to , the sum is zero because

the positives and negatives all cancel out. Start with the next number:

The simplest way to handle this question is to consider the sequence as being a set of 3 repeating terms, (3,2,4). Within the first 18 terms in the sequence, this pattern will repeat a total of 6 times.

The sum of one repeat is:

Since there are 6 repeats, take the sum of one repeat (9) and multiply by the number of repeats:

The sum of the first 18 terms is 54.