In this geometric series, $\left\{ {4, - 8,16, - 32,...} \right\}$. What is the value of the ”$q$” term?

Hint: Generally, a geometric sequence is a sequence in which a term is a multiple of the previous term, and a geometric sequence is an infinite sequence. And a geometric series is the sum of an infinite number of terms. Here, we need to find the ${q^{th}}$term of $\left\{ {4, - 8,16, - 32,...} \right\}$. To find this we have to find the first term and common ratio. Then we need to just apply the below formula.

Formula to be used:
The ${n^{th}}$ term of GP is as follows.
${T_n} = a{r^{n - 1}}$
Here ${T_n}$ is the ${n^{th}}$ term of GP, $a$ is the first term, and $r$is the common ratio.

Complete answer:
We denote the geometric series in general form as $a + ar + a{r^2} + a{r^3} + ...$
Here, the most important parameters are: $a$ is the first term and $r$is the common ratio.
The first term is the coefficient of each term and each term has the same coefficient.
Also, the common ratio is the ratio of any term with the previous term in the given series.
The given sequence is $\left\{ {4, - 8,16, - 32,...} \right\}$
Here, $a = 4$ and $ar = - 8$.
Thus, $\dfrac{{ar}}{a} = \dfrac{{ - 8}}{4}$
$ \Rightarrow r = - 2$
We need to find the ${q^{th}}$ term of the series.
We know that the ${n^{th}}$ term of GP is ${T_n} = a{r^{n - 1}}$.
Thus, ${T_q} = a{r^{q - 1}}$
$ \Rightarrow {T_q} = 4 \times {\left( { - 2} \right)^{q - 1}}$
Thus, $4 \times {\left( { - 2} \right)^{q - 1}}$ is the required ${q^{th}}$ term of the series.

Note: The very simple method to find the ${n^{th}}$ term of the series is found by using the formula ${T_n} = a{r^{n - 1}}$. We can also find the ${n^{th}}$ term of the series without using the formula. First, we need to note the multiple of the previous term for each term and it is the same for all terms. The same step is to be followed for the ${n^{th}}$ term of the series. But this method will be quite confusing. So we shall follow the formula.