This sum can be determined using the formula for the sum of an infinite geometric series, with initial term $\displaystyle a_{0} = 1$ and common ratio $\displaystyle r = -\frac{2}{3}$:

$\displaystyle S = \frac{a_{0}} {1-r} = \frac{1} {1- \left ( -\frac{2}{3}\right ) } = \frac{1} {1+ \frac{2}{3} }= \frac{1} { \frac{5}{3} } =\frac{3}{5}$

An arithmetic sequence is one in which there is a common difference between consecutive terms. For example, the sequence {2, 5, 8, 11} is an arithmetic sequence, because each term can be found by adding three to the term before it. 

Let $\displaystyle a_{n}$ denote the nth term of the sequence. Then the following formula can be used for arithmetic sequences in general:

$\displaystyle a_n=a_1 +(n-1)d$, where d is the common difference between two consecutive terms. 

We are given the 4th and 8th terms in the sequence, so we can write the following equations:

$\displaystyle a_4=a_1+(4-1)d=a_1+3d=-20$

$\displaystyle a_8=a_1+(8-1)d=a_1+7d=-10$.

We now have a system of two equations with two unknowns:

$\displaystyle a_1+3d=-20$

$\displaystyle a_1+7d=-10$

Let us solve this system by subtracting the equation $\displaystyle a_1+7d=-10$ from the equation $\displaystyle a_1+3d=-20$. The result of this subtraction is

$\displaystyle -4d=-10$.

This means that d = 2.5.

Using the equation $\displaystyle a_1+7d=-10$, we can find the first term of the sequence.

$\displaystyle a_1+7(2.5)=-10$

$\displaystyle a_1=-27.5$

Ultimately, we are asked to find the hundredth term of the sequence.

$\displaystyle a_{100}=a_1+(100-1)d=-27.5+99(2.5)=220$

The answer is 220.

The sum of all odd numbers is another way to construct perfect squares.  To see why this is, we can construct the series as follows.

$\displaystyle S_k = 1 + 3 + 5 + \cdots (2k+1)$

We draw $\displaystyle k$ from the series by subtracting one from each term.

$\displaystyle S_k = k + 0 + 2 + 4 + \cdots + 2k$

We discard the 0 term and factor 2 out of the remaining terms.

$\displaystyle S_k = k + 2(1 + 2 + 3 + \cdots +k)$

And finally we use the property that $\displaystyle \sum\limits_{i=1}^n i=\frac{n(n-1)}{2}$ to evaluate the series.  

$\displaystyle S_k = k + 2*\frac{k(k-1)}{2} = k + k^2-k = k^2$

The smallest value of $\displaystyle k$ where the square exceeds 200 is $\displaystyle k=15$.

The terms of an arithmetic series are generated by the relation

$\displaystyle a_{n}=a_{1}+(n-1)d$,

where $\displaystyle a_{1}$ is the 1st term, $\displaystyle a_{n}$ is the nth term, and d is the common difference.

For $\displaystyle n=1$

$\displaystyle a_{1}=3$,

for $\displaystyle n=9$

$\displaystyle a_{9}=35$.

The first step is to find $\displaystyle d$.

$\displaystyle 35=3+(9-1)d$

$\displaystyle 32=8d$, so

$\displaystyle d=32/8=4$.

Now to find $\displaystyle a_{n}$, when $\displaystyle n=17$.

Use the generating relation

$\displaystyle a_{17}=3+(17-1)\cdot 4=3+16\cdot 4=3+64=67$.

To find the next term, we need to figure out what is happening from one term to the next.

From 2 to 5, we can see that 3 is added.

From 5 to 14, 9 is added.

From 14 to 41, 27 is added. 

If you look closely, you can notice a trend.

The amount added each time triples. Therefore, the next amount added should be

$\displaystyle 3 \cdot 27 =81$

Thus, $\displaystyle 41 + 81 = 122$

In the series given, $\displaystyle 4$ is added to each previous term to get the next term. Since a fixed number is ADDED each time, this series can be categorized as an arithmetic series. 

First, we need to figure out what the pattern is in this series. Notice how each term results in the following term. In this case, each term is multiplied by $\displaystyle 3$ to get the next term. Since each term is MULTIPLIED by a fixed number, this can be defined as a geometric series. 

Common ratio is the number that is multiplied by each term to get the next term in a geometric series. Since the first two terms are $\displaystyle 5$ and $\displaystyle 10$, we look at what is multiplied between these. Once way to determine this if not immediately obvious is to divide the second term by the first term. In this case we get: 

$\displaystyle 10\div5 = 2$ which gives us our common ratio. 

This equation is a series in summation notation.

$\displaystyle \sum_{k=3}^{8}{\frac{1}{k}}$

We can see that the bottom k=3 designates where the series starts, 8 represents the stop point, and 1/k represents the rule for summation. We can expand this equation as follows:

$\displaystyle \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}$

Here, we have just substituted "k" for each value from 3 to 8. To solve, we must then find the least common demoninator. That would be 280. This can be found in several ways, such as separating the fractions by like denominators:

$\displaystyle \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$

$\displaystyle \frac{2}{8} + \frac{4}{8} + \frac{1}{5} + \frac{1}{7} = \frac{306}{8(7)(5)} =\frac{153}{140}$

We note that there is no common difference between $\displaystyle a_n$ and $\displaystyle a_{n+1}$ so the sequence cannot be arithmetic.

We also note that there exists a common ratio between two consecutive terms.

$\displaystyle \frac{a_{n+1}}{a_n}=2$

Since there exists a common ratio, the sequence is Geometric.