Arithmetic Sequences - GRE Quantitative Reasoning

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.

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:

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:

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.

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:

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:

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.

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:

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:

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:

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:

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 less 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 th element, you will have:

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:

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:

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 less 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 th element, you will have:

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:

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:

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 less 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 th element, you will have:

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.

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