The formula for the sum of a geometric series is

$\displaystyle S=\frac{a_1-a_1r^n}{1-r}$,

where $\displaystyle a_1$ is the first term in the series, $\displaystyle r$ is the rate of change between sequential terms, and $\displaystyle n$ is the number of terms in the series

For this problem, these values are:

$\displaystyle a_1 = \frac{1}{4}$

$\displaystyle r = 2$

$\displaystyle n = 12$

Plugging in our values, we get:

$\displaystyle S=\frac{\frac{1}{4}-\frac{1}{4}(2)^{12}}{1-2}$

$\displaystyle S = 1023.75$

The formula for the sum of a geometric series is

$\displaystyle S=\frac{a_1-a_1r^n}{1-r}$,

where $\displaystyle a_1$ is the first term in the series, $\displaystyle r$ is the rate of change between sequential terms, and $\displaystyle n$ is the number of terms in the series

In this problem we have:

$\displaystyle a_1 = \frac{1}{2}$

$\displaystyle r = 4$

$\displaystyle n = 10$

Plugging in our values, we get:

$\displaystyle S=\frac{\frac{1}{2}-\frac{1}{2}(4)^{10}}{1-4}$

$\displaystyle S = 174,762.5$

The sequence can be described by the equation $\displaystyle \small 2+3(n-1)$, where $\displaystyle n$ is the term in the sequence.

For the 15th term, $\displaystyle n=15$.

$\displaystyle \small 2+3(n-1)=2+3(15-1)$

$\displaystyle 2+3(14)$

$\displaystyle 2+42$

$\displaystyle 44$

To find the first three terms, replace $\displaystyle n$ with $\displaystyle 1$, $\displaystyle 2$, and $\displaystyle 3$.

$\displaystyle (4n-5) = (4(1)-5)=-1$

$\displaystyle (4n-5) = (4(2)-5)=3$

$\displaystyle (4n-5) = (4(3)-5)=7$

The first three terms are $\displaystyle -1$, $\displaystyle 3$, and $\displaystyle 7$.

To find the first three terms, replace $\displaystyle n$ with $\displaystyle 3$, $\displaystyle 4$, and $\displaystyle 5$.

$\displaystyle (8-5n) = (8-5(3))=-7$

$\displaystyle (8-5n) = (8-5(4))=-12$

$\displaystyle (8-5n) = (8-5(5))=-17$

The first three terms are $\displaystyle -7$, $\displaystyle -12$, and $\displaystyle -17$.

In the arithmetic series, the first terms can be found by plugging $\displaystyle 1$, $\displaystyle 2$, and $\displaystyle 3$ into the equation.

$\displaystyle x=1$

$\displaystyle 4(1)-5 = -1$

$\displaystyle x=2$

$\displaystyle 4(2)-5 = 3$

$\displaystyle x=3$

$\displaystyle 4(3)-5 = 7$

In the arithmetic series, the first terms can be found by plugging in $\displaystyle 3$, $\displaystyle 4$, and $\displaystyle 5$ for $\displaystyle c$.

$\displaystyle c=3$

$\displaystyle 8-5(3) = -7$

$\displaystyle c=4$

$\displaystyle 8-5(4) = -12$

$\displaystyle c=5$

$\displaystyle 8-5(5) = -17$

The first terms can be found by substituting $\displaystyle -2$, $\displaystyle -1$, and $\displaystyle 0$ for $\displaystyle m$ into the sum formula.

$\displaystyle m=-2$

$\displaystyle (2)^{-2}=\frac{1}{4}$

$\displaystyle m=-1$

$\displaystyle (2)^{-1}=\frac{1}{2}$

$\displaystyle m=0$

$\displaystyle (2)^{0}=1$

The first terms can be found by substituting $\displaystyle 2$, $\displaystyle 3$, and $\displaystyle 4$ in for $\displaystyle b$.

$\displaystyle b=2$

$\displaystyle \frac{1}{2}(4)^{2-2}=\frac{1}{2}$

$\displaystyle b=3$

$\displaystyle \frac{1}{2}(4)^{3-2}=2$

$\displaystyle b=4$

$\displaystyle \frac{1}{2}(4)^{4-2}=8$

We will need to use the Binomial Theorem in order to solve this problem. Consider the expansion of $\displaystyle (a+b)^n$, where n is an integer. The rth term of this expansion is given by the following formula:

$\displaystyle \binom{n}{r-1}a^{n-r+1}b^{r-1}$,

 where $\displaystyle \binom{n}{r-1}$ is a combination. In general, if x and y are nonnegative integers such that x > y, then the combination of x and y is defined as follows: $\displaystyle \binom{x}{y}=_xC_y=\frac{x!}{(y)!(x-y)!}$.

We are asked to find the sixth term of $\displaystyle \left (\frac{1}{2}x-2 \right )^{10}$, which means that in this case r = 6 and n = 10. Also, we will let $\displaystyle a=\frac{1}{2}x$ and $\displaystyle b=-2$. We can now apply the Binomial Theorem to determine the sixth term, which is as follows:

$\displaystyle \binom{10}{6-1}\left (\frac{1}{2}x \right )^{10-6+1}\cdot(-2) ^{6-1}$

$\displaystyle =\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$

Next, let's find the value of $\displaystyle \binom{10}{5}$. According to the definition of a combination, 

$\displaystyle \binom{10}{5}=\frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5!}{5!5!}$

$\displaystyle =\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=252$.

Remember that, if n is a positive integer, then $\displaystyle n!=n\cdot(n-1)\cdot(n-2)\cdot\cdot\cdot3\cdot2\cdot1$. This is called a factorial. 

Let's go back to simplifying $\displaystyle \binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$.

 $\displaystyle \binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}=252\cdot\left (\frac{1}{2}x \right )^{5}\cdot-32$

$\displaystyle =252\cdot\left (\frac{1}{2} \right )^5\cdot x^5\cdot (-32) =252\cdot \frac{1}{32}\cdot-3 2\cdot x^5$

$\displaystyle =-252x^5$

The answer is $\displaystyle -252x^5$.