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^th element, continue following this:

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

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

The first term of the sequence is \(\displaystyle \small a_{1}\), so here \(\displaystyle \small a_{1} = 3\), 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.

\(\displaystyle a_1+2n+n^2\)

\(\displaystyle \small a_{1} = 3\) and \(\displaystyle n=20\)

\(\displaystyle \small 3 + 2(20) + 20^{2} = 443\)

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:

\(\displaystyle z_{n+1} = 3z_n-7\)

Consequently,

\(\displaystyle z_n=\frac{z_{n+1}+7}{3}\)

Using the first equation, we can define \(\displaystyle z_5\) in terms of \(\displaystyle z_3\):

\(\displaystyle z_5=3z_4-7=3(3z_3-7)-7=9z_3-21-7=9z_3-28\)

This allows us to rewrite

\(\displaystyle z_3+z_5=142\)

as

\(\displaystyle z_3+9z_3-28=142\)

Rearrangement of terms allows us to solve for \(\displaystyle z_3\):

\(\displaystyle 10z_3=170\)

\(\displaystyle z_3=17\)

Now, using our second equation, we can find \(\displaystyle z_1\), the first term:

\(\displaystyle z_{1}=\frac{z_2+7}{3}=\frac{\frac{z_{3}+7}{3}+7}{3}=\frac{\frac{24}{3}+7}{3}=\frac{15}{3}=5\)

Begin by interpreting the general definition:

\(\displaystyle s_n=s_n_-_1 + 5\)

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

\(\displaystyle s_5=s_4+5\)

It is easiest to count upwards:

\(\displaystyle s_2=14+5=19\)

\(\displaystyle s_3=19+5=24\)

\(\displaystyle s_4=24+5=29\)

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:

\(\displaystyle s_n=s_n_-_1 + 11\)

This means that for every element in the list, each one is \(\displaystyle 11\) greater than the one preceding it.  For instance:

\(\displaystyle s_1_0_1=s_1_0_0+11\)

Now, notice that the first element is:

\(\displaystyle 201\)

The second is:

\(\displaystyle 201+11\)

The third could be represented as:

\(\displaystyle 201+11+11\)

And so forth...

Now, notice that for the third element, there are only two instances of \(\displaystyle 11\).  We could rewrite our sequence:

\(\displaystyle 201+2(11)\)

This value will always "lag behind" by one.  Therefore, for the \(\displaystyle 101\)st element, you will have:

\(\displaystyle 201+100\cdot11=1301\)

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:

\(\displaystyle s_n=s_n_-_1 -21\)

This means that for every element in the list, each one is \(\displaystyle 21\) less than the one preceding it.  For instance:

\(\displaystyle s_5_7=s_5_6-21\)

Now, notice that the first element is:

\(\displaystyle 93\)

The second is:

\(\displaystyle 93-21\)

The third could be represented as:

\(\displaystyle 93-21-21\)

And so forth...

Now, notice that for the third element, there are only two instances of \(\displaystyle -21\).  We could rewrite our sequence:

\(\displaystyle 93-2*21\)

This value will always "lag behind" by one.  Therefore, for the \(\displaystyle 57\)th element, you will have:

\(\displaystyle 93-56\cdot21=-1083\)