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 \(\displaystyle -252x^5\).