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.

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^thelement, continue following this:

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

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

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

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

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

Consequently,

Using the first equation, we can definein 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:

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 isgreater than the one preceding it. For instance:

Now, notice that the first element is:

The second is:

The third could be represented as:

一个nd so forth...

Now, notice that for the third element, there are onlytwoinstances of. We could rewrite our sequence:

This value will always "lag behind" by one. Therefore, for thest 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 isless than the one preceding it. For instance:

Now, notice that the first element is:

The second is:

The third could be represented as:

一个nd so forth...

Now, notice that for the third element, there are onlytwoinstances of. We could rewrite our sequence:

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