Question

Write down the first five terms of the geometric progression which has first term 1 and common ratio 4.
A) 1, 4, 16, 64, 244
B) 1, 4, 24, 64, 256
C) 1, 4, 16, 32, 256
D) 1, 4, 16, 64, 256

Answer 1

Here in the solution, we will use the concept and formula of geometric progression. Geometric progression is a sequence of terms where the subsequent term is found by multiplying the previous term with a constant. This constant is known as the common ratio. We will find the GP series, by substituting the values in the formula to get the subsequent terms.

Formula used:
Here, we will be using the formula of a geometric progression, ${a_n} = a{r^{n - 1}}$, where ${a_n}$ denotes the ${n^{{\rm{th}}}}$ term,
$a$ denotes the first term, $r$ denotes the common ratio, and $n$ is the number of terms.

Complete step-by-step answer:
In this question, we are required to find the geometric progression series. To do this, we will use the formula for the geometric progression.
${a_n} = a{r^{n - 1}}$
We will substitute the different values in this formula to obtain the series.
We are given that the common ratio is 4. This means that $r = 4$.
Also, we are given that the first term is 1. This means that $a = 1$.
As we already know the first term, let us begin by finding the second term now.
Substituting $a = 1$, $r = 4$, and $n = 2$ in the formula ${a_n} = a{r^{n - 1}}$, we get
${a_2} = 1 \cdot {4^{2 - 1}}$
Subtracting the terms of the exponent, we get
${a_2} = 1 \cdot {4^1}$
Now, simplifying the above expression, we get
${a_2} = 4$
Let us find the third term now. We will substitute $a = 1$, $r = 4$, and $n = 3$ in the formula ${a_n} = a{r^{n - 1}}$. ${a_3} = 1 \cdot {4^{3 - 1}}$
Subtracting the terms of the exponent, we get
${a_3} = 1 \cdot {4^2}$
We will now apply the exponent on the term 4.
${a_3} = 1 \cdot 16$
Now, multiplying the terms, we get
${a_3} = 16$
Let us find the fourth term now. Substitute $a = 1$, $r = 4$, and $n = 4$ in the formula ${a_n} = a{r^{n - 1}}$. Solve the obtained expression using the basic arithmetic to obtain the value of ${a_4}$.
$\begin{array}{c}{a_4} = 1 \cdot {4^{4 - 1}}\\\ = 1 \cdot {4^3}\\\ = 1 \cdot 64\\\ = 64\end{array}$
Let us find the fifth term now. Substitute $a = 1$, $r = 4$, and $n = 5$ in the formula ${a_n} = a{r^{n - 1}}$. Solve the obtained expression using the basic arithmetic to obtain the value of ${a_5}$.
$\begin{array}{c}{a_5} = 1 \cdot {4^{5 - 1}}\\\ = 1 \cdot {4^5}\\\ = 1 \cdot 256\\\ = 256\end{array}$
Thus, the first five numbers in the given geometric progression series are $1,4,16,64,256...$.

Hence, the correct option is D.

Note: There is an alternative way too to solve this question. We know that the geometric progression series is obtained by multiplying every preceding term with a constant number that is known as the common ratio. This gives every new term. Thus, based on this concept the given geometric series can be determined as follows, when we know that the first term is 1 and the common ratio is 4.
${a_1} = 1$ (given)
To find the second term, multiply the first term with the common ratio.
$\begin{array}{c}{a_2} = {a_1} \cdot 4\\\ = 1 \cdot 4\\\ = 4\end{array}$
To find the third term, multiply the second term with the common ratio.
$\begin{array}{c}{a_3} = {a_2} \cdot 4\\\ = 4 \cdot 4\\\ = 16\end{array}$
To find the fourth term, multiply the third term with the common ratio.
$\begin{array}{c}{a_4} = {a_3} \cdot 4\\\ = 16 \cdot 4\\\ = 64\end{array}$
Lastly, to find the fifth term, multiply the fourth term with the common ratio.
$\begin{array}{c}{a_5} = {a_4} \cdot 4\\\ = 64 \cdot 4\\\ = 256\end{array}$
Thus, the first five numbers of the geometric progression series are $1,4,16,64,256...$.