### All GRE Math Resources

## Example Questions

### Example Question #1 : Sequences

What is the sum of the odd integers ?

**Possible Answers:**

None of the other answers

**Correct answer:**

Do **NOT** try to add all of these. It is key that you notice the pattern. Begin by looking at the first and the last elements: 1 and 99. They add up to 100. Now, consider 3 and 97. Just as 1 + 99 = 100, 3 + 97 = 100. This holds true for the entire list. Therefore, it is crucial that we find the number of such pairings.

1, 3, 5, 7, and 9 are paired with 99, 97, 95, 93, and 91, respectively. Therefore, for each 10s digit, there are 5 pairings, or a total of 500. To get all the way through our numbers, you will have to repeat this process for the 10s, 20s, 30s, and 40s (all the way to 49 + 51 = 100).

Therefore, there are 500 (per pairing) * 5 pairings = 2500.

### Example Question #2 : Sequences

A sequence is defined by the following formula:

What is the 4th element of this sequence?

**Possible Answers:**

**Correct answer:**

With series, you can always "walk through" the values to find your answer. Based on our equation, we can rewrite as :

You then continue for the third and the fourth element:

### Example Question #1 : Sequences

What is the sum of the 40th and the 70th elements of the series defined as:

**Possible Answers:**

**Correct answer:**

When you are asked to find elements in a series that are far into its iteration, you need to find the pattern. You absolutely cannot waste your time trying to calculate all of the values between and . Notice that for every element after the first one, you subtract . Thus, for the second element you have:

For the third, you have:

Therefore, for the 40th and 70th elements, you will have:

The sum of these two elements is:

### Example Question #1 : Sequences

The first term in a sequence of integers is 2 and the second term is 10. All subsequent terms are the arithmetic mean of all of the preceding terms. What is the 39th term?

**Possible Answers:**

600

6

5

300

1200

**Correct answer:**

6

The first term and second term average out to 6. So the third term is 6. Now add 6 to the preceding two terms and divide by 3 to get the average of the first three terms, which is the value of the 4th term. This, too, is 6 (18/3)—all terms after the 2nd are 6, including the 39th. Thus, the answer is 6.

### Example Question #1 : Sequences

Consider the following sequence of integers:

5, 11, 23, 47

What is the 6^{th} element in this sequence?

**Possible Answers:**

95

93

189

191

None of the other answers

**Correct answer:**

191

First, consider the change in each element. Notice that in each case, a given element is twice the preceding one plus one:

11 = 2 * 5 + 1

23 = 11 * 2 + 1

47 = 23 * 2 + 1

To find the 6^{th} element, continue following this:

The 5^{th}: 47 * 2 + 1 = 95

The 6^{th}: 95 * 2 + 1 = 191

### Example Question #1 : Sequences

The sequence begins with the numbers and has the term defined as , for .

What is the value of the term of the sequence?

**Possible Answers:**

**Correct answer:**

The first term of the sequence is , so here , and we're interested in finding the 20th term, so we'll use n = 20.

Plugging these values into the given expression for the nth term gives us our answer.

and

### Example Question #1 : Sequences

In a sequence of numbers, the first two values are 1 and 2. Each successive integer is calculated by adding the previous two and mutliplying that result by 3. What is fifth value in this sequence?

**Possible Answers:**

None of the other answers

**Correct answer:**

Our sequence begins as 1, 2.

Element 3: (Element 1 + Element 2) * 3 = (1 + 2) * 3 = 3 * 3 = 9

Element 4: (Element 2 + Element 3) * 3 = (2 + 9) * 3 = 11 * 3 = 33

Element 5: (Element 3 + Element 4) * 3 = (9 + 33) * 3 = 42 * 3 = 126

### Example Question #5 : Sequences

Let Z represent a sequence of numbers wherein each term is defined as seven less than three times the preceding term. If , what is the first term in the sequence?

**Possible Answers:**

**Correct answer:**

Let us first write the value of a consecutive term in a numerical format:

Consequently,

Using the first equation, we can define in terms of :

This allows us to rewrite

as

Rearrangement of terms allows us to solve for :

Now, using our second equation, we can find , the first term:

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

The sequence is defined by:

What is ?

**Possible Answers:**

**Correct answer:**

Begin by interpreting the general definition:

This means that every number in the sequence is five greater than the element preceding it. For instance:

It is easiest to count upwards:

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

The sequence is defined by:

What is the value of ?

**Possible Answers:**

**Correct answer:**

For this problem, you definitely do not want to "count upwards" to the full value of the sequence. Therefore, the best approach is to consider the general pattern that arises from the general definition:

This means that for every element in the list, each one is greater than the one preceding it. For instance:

Now, notice that the first element is:

The second is:

The third could be represented as:

And so forth...

Now, notice that for the third element, there are only *two* instances of . We could rewrite our sequence:

This value will always "lag behind" by one. Therefore, for the st element, you will have:

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