Chapter 13, Problem 1PCT

In Problems 1–6, determine if the sequence is arithmetic, geometric, or neither. If arithmetic or geometric, determine the first term and the common difference or common ratio.

1. $\frac{3}{4},\frac{3}{16},\frac{3}{64},\frac{3}{256},\cdots$

To determine

Whether the given sequence is arithmetic, geometric or neither and find the first term $a_1$ and common difference or common ratio if it is arithmetic or geometric.

Answer to Problem 1PCT

The sequence $\frac{3}{4},\frac{3}{16},\frac{3}{64},\frac{3}{256},\mathrm{...}$ is geometric sequence. The first term is $a_1=\frac{3}{4}$ and the common ratio is $r=\frac{1}{4}$.

Explanation of Solution

Definition used:

Arithmetic:

“If the difference between consecutive terms is always the same number, then the sequnce is arithmetic sequence.

If the first term is called $a_1$ and the common difference between the consecutive terms is called d, then the terms the pattern of an arithmetic sequence is $a_1,a_1+d,a_1+2d,\mathrm{...}$”.

Geometric:

“If the ratio of consecutive terms in a sequence is constant, then the sequnce is a geometric sequence.

If the first term is called $a_1$ and the common ratio of consecutive terms is called r, then the pattern of a geometric sequence is $a_1,a_{1}r,a_1{r}^2,a_1{r}^3,\mathrm{...}$”.

Calculation:

From the given sequence, the first, second, third and fourth term are $a_1=\frac{3}{4},a_2=\frac{3}{16},a_3=\frac{3}{64}\text{ and }{a}_4=\frac{3}{256}$, respectively.

Obtain the difference between the consecutive terms.

Compute $a_2-a_1$.

$\begin{array}{c}{a}_2-a_1=\frac{3}{16}-\frac{3}{4}\\ =\frac{3}{16}-\frac{12}{16}\\ =\frac{3-12}{16}\\ =-\frac{9}{16}\end{array}$

Compute $a_3-a_2$.

$\begin{array}{c}{a}_3-a_2=\frac{3}{64}-\frac{3}{16}\\ =\frac{3}{64}-\frac{12}{64}\\ =\frac{3-12}{64}\\ =-\frac{9}{64}\end{array}$

Compute $a_4-a_3$.

$\begin{array}{c}{a}_4-a_3=\frac{3}{256}-\frac{3}{64}\\ =\frac{3}{256}-\frac{12}{256}\\ =\frac{3-12}{256}\\ =-\frac{9}{256}\end{array}$

Since the difference between the consecutive terms is not constant, the given sequence is not arithmetic.

Therefore, the sequence $\frac{3}{4},\frac{3}{16},\frac{3}{64},\frac{3}{256},\mathrm{...}$ is not arithmetic sequence.

Obtain the ratio of the consecutive terms.

Compute $\frac{a_2}{a_1}$.

$\begin{array}{c}\frac{a_2}{a_1}=\frac{\frac{3}{16}}{\frac{3}{4}}\\ =\frac{3}{16}\cdot \frac{4}{3}\\ =\frac{1}{4}\end{array}$

Compute $\frac{a_3}{a_2}$.

$\begin{array}{c}\frac{a_3}{a_2}=\frac{\frac{3}{64}}{\frac{3}{16}}\\ =\frac{3}{64}\cdot \frac{16}{3}\\ =\frac{1}{4}\end{array}$

Compute $\frac{a_4}{a_3}$.

$\begin{array}{c}\frac{a_4}{a_3}=\frac{\frac{3}{256}}{\frac{3}{64}}\\ =\frac{3}{256}\cdot \frac{64}{3}\\ =\frac{1}{4}\end{array}$

Clearly, the common ratio is $\frac{1}{4}$.

Since the ratio of the consecutive terms is constant, the given sequence is geometric.

Therefore, the sequence $\frac{3}{4},\frac{3}{16},\frac{3}{64},\frac{3}{256},\mathrm{...}$ is geometric sequence. The first term is $a_1=\frac{3}{4}$ and the common ratio is $r=\frac{1}{4}$.