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 $\displaystyle s_2$ as :

$\displaystyle s_2 = 3\cdot s_1+5=3\cdot 12+5=36+5 = 41$

You then continue for the third and the fourth element:

$\displaystyle s_3=3\cdot 41 + 5=128$

$\displaystyle s_4=3\cdot 128+5=389$

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 $\displaystyle 1$ and $\displaystyle 70$. Notice that for every element after the first one, you subtract $\displaystyle 5$. Thus, for the second element you have:

$\displaystyle s_2=281-5$

For the third, you have:

$\displaystyle s_3 = 281 - (2)5$

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

$\displaystyle s_4_0 = 281 - (39)5 = 281 - 195 = 86$

$\displaystyle s_7_0 = 281 - (69)5 = 281-345=-64$

The sum of these two elements is:

$\displaystyle 86-64=22$

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:

$\displaystyle 201+100\cdot11=1301$