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.

Let's call the first term in the sequence a1 and the nth term an.

We are told that each term is r times larger than the one before it. Thus, we can find the next term in the sequence by multiplying by r.

a1 = a1

a2 = r(a1)

a3 = r(a2) = r(r(a1)) = r2(a1)

a4 = r(a3) = r(r2(a1)) = r3(a1)

an = r(n–1)a1

We can use this information to find r.

The problem gives us the value of the third and the sixth terms.

a3 = r2(a1) = 12

a6 = r5(a1) = 96

Let's solve for a1 in terms of r and a3.

a1 = 12/(r2)

Let's then solve for a1 in terms of r and a6.

a1 = 96/(r5)

Now, we can set both values equal and solve for r.

12/(r2) = 96/(r5)

Multiply both sides by r5 to get rid of the fraction.

12r5/r2 = 96

Apply the property of exponents which states that ab/ac = ab–c.

12r3 = 96

Divide by 12 on both sides.

r3 = 8

Take the cube root of both sides.

r = 2

This means that each term is two times larger than the one before it, or that each term is one half as large as the one after it.

a2 must equal a3 divided by 2, which equals 12/2 = 6.

a1 must equal a2 divided by 2, which equals 6/2 = 3.

Here are the first eight terms of the sequence:

3, 6, 12, 24, 48, 96, 192, 384

The question asks us to find the sum of all the terms less than 250. Only the first seven terms are less than 250. Thus the sum is equal to the following:

sum = 3 + 6 + 12 + 24 + 48 + 96 + 192 = 381

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

Given terms of an arithmetic sequence, , the common difference is

Setting , the common difference is found to be

The term of the sequence is

Setting , this term is found to be

It is not necessary to determine the common difference or any other terms to answer this question. If we let be the common difference, then

and

, or

The sum of the second and ninth terms is given by the expression

Given terms of an arithmetic sequence, , the common difference is

.

Set :

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.

Each term in the sequence is one less than twice the previous term.

So,

Notice that in the sequence

each term increases by .

It is always good strategy when attempting to find a pattern in a sequence to examine the difference between each term.

We continue the pattern to find:

The term is

The term is

The term is

It is useful to note that the sequence is defined by,

where n is the number of any one term.

We can solve

to find the term.

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.

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.

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.