A geometric sequence is one in which the next term is found by mutlplying the previous term by a particular constant. Thus, we look for an implicit definition which involves multiplication of the previous term. The only possibility is: 

\(\displaystyle t_{0}=4\)

\(\displaystyle t_{n}=t_{n-1}\times 3\)

This is a geometric series. The explicit formula for any geometric series is:

\(\displaystyle a_{n}= a_{1} \cdot r^{n-1}\), where \(\displaystyle r\) is the common ratio and \(\displaystyle n\) is the number of terms.

In this instance \(\displaystyle r = -2\) and \(\displaystyle a_1=-1\).

\(\displaystyle a_{n}= a_{1} \cdot r^{n-1}\rightarrow a_n=(-1)\cdot(-2)^{n-1}\)

Substitute \(\displaystyle n = 20\) into the equation to find the 20th term:

\(\displaystyle a_{20}=(-1)\cdot(-2)^{20-1}\)

\(\displaystyle a_{20}=-(-2)^{19}=-(-524288)=524288\)

This series is neither geometric nor arithmetic. 

A geometric sequences is multiplied by a common ratio (\(\displaystyle r\)) each term.  An arithmetic series adds the same additional amount (\(\displaystyle d\)) to each term.  This series does neither.

Mutiplicative and subtractive are not types of sequences.

Therefore, the answer is none of the other answers.

The explicit formula for a geometric series is \(\displaystyle a_{n}=a_{1}\cdot r^{n-1}\)

In this problem \(\displaystyle r=\frac{1}{2}\)

Therefore:

\(\displaystyle a_{10}=\frac{1}{4}\cdot \frac{1}{2}^{9}\)

\(\displaystyle {\color{Red} a_{10}=\frac{1}{2048}}\)

The explicit formula for a geometric series is \(\displaystyle a_n=a_1\cdot r^{n-1}\).

Therefore, \(\displaystyle {\color{Red} a_6=3\cdot 7^5, a_1=3}\) is the only answer that works.

This series is geometric.  The explicit formula for any geometric series is:

\(\displaystyle a_n=a_1\cdot r^{n-1}\)

Where \(\displaystyle n\) represents the \(\displaystyle n^{th}\) term, \(\displaystyle a_1\) is the first term, and \(\displaystyle r\) is the common ratio.

In this series \(\displaystyle r = \frac{1}{2}\). 

Therefore the formula to find the 15th term is:

\(\displaystyle a_{15}=120\cdot \left(\frac{1}{2}\right)^{14}\)

\(\displaystyle a_{15}=.0073\)

\(\displaystyle \textup{Look for a pattern. Each term in the sequence is three times the previous term.}\)

\(\displaystyle 18\ast3=54\)

First we need to find the common ratio by dividing the second term by the first: 

\(\displaystyle -3 \div 2 = -1.5\)

The \(\displaystyle n^{th}\) term is

\(\displaystyle \\a_{n} = a_1(r)^{n-1}\\a_n=2(-1.5)^{n-1}\),

so the 33rd term will be

\(\displaystyle a_{33} = 2(-1.5)^ {32 } = 862,879.767\).

First find the common ratio by dividing the second term by the first:

\(\displaystyle 2,800 \div 7,000 = \frac{ 2}{5} = 0.4\)

Since the first term is \(\displaystyle 7,000\), the nth term can be found using the formula

\(\displaystyle \\a_{n} = a_1(r)^{n-1}\) 

\(\displaystyle 7,000(0.4)^ {n-1 }\),

so the 19th term is \(\displaystyle 7,000(0.4)^{18} = 0.000481\)

First, find the common ratio by dividing the second term by the first:

\(\displaystyle 108 \div 90 = 1.2\)

The nth term can be found using

\(\displaystyle \\a_{n} = a_1(r)^{n-1}\)

\(\displaystyle 90(1.2)^ {n-1}\),

so the 21st term is

\(\displaystyle 90(1.2) ^ {20 } = 3,450.384\).