Chapter 5 - Number Theory and the Real Number System - Chapter Summary, Review, and Test - Review Exercises - Page 337: 145

The given sequence is a geometric sequenceand the next two terms for the given sequence are\[\frac{1}{256}\ \text{and }\frac{1}{1024}\].

To check is a sequence is an arithmetic sequence, see if the differences between two consecutive terms are equal. That is, check if {a}_{n+1}-{a}_{n}=d for all n in N, here \[d\] is the common difference. To check if a sequence is a geometric sequence, see if the ratio between two consecutive terms is equal. That is, check if {{a}_{n+1}}/{{a}_{n}}=r for all n in Nu, here \[r\] is the common ratio. When\[n=1\], for the given sequence, \[\begin{align} & d={{a}_{2}}-{{a}_{1}} \\ & =\frac{1}{4}-1 \\ & =-\frac{3}{4} \\ & =-0.75 \end{align}\] And, \[\begin{align} & r=\frac{{{a}_{2}}}{{{a}_{1}}} \\ & =\frac{1}{4}\div 1 \\ & =\frac{1}{4} \end{align}\] When\[n=2\], for the given sequence, \[\begin{align} & d={{a}_{3}}-{{a}_{2}} \\ & =\frac{1}{16}-\frac{1}{4} \\ & =-\frac{3}{16} \\ & =-0.18 \end{align}\] And, \[\begin{align} & r=\frac{{{a}_{3}}}{{{a}_{2}}} \\ & =\frac{1}{16}\div \frac{1}{4} \\ & =\frac{1}{4} \end{align}\] When\[n=3\], for the given sequence, \[\begin{align} & d={{a}_{4}}-{{a}_{3}} \\ & =\frac{1}{64}-\frac{1}{16} \\ & =-\frac{3}{64} \\ & =-0.04 \end{align}\] And, \[\begin{align} & r=\frac{{{a}_{4}}}{{{a}_{3}}} \\ & =\frac{1}{64}\div \frac{1}{16} \\ & =\frac{1}{4} \end{align}\] Since, \[r=\frac{1}{4}\ \ \forall n=1,2,3\]and \[d\]is not equal for all n in N, it implies the given sequence is a geometric sequence. Use the formula \[{{a}_{n}}={{a}_{1}}{{r}^{n-1}}\]to find the next term of the arithmetic sequence. Put \[n=5\]in\[{{a}_{n}}={{a}_{1}}{{r}^{n-1}}\]to find the fifth term of the given arithmetic sequence as follows: \[\begin{align} & {{a}_{5}}={{a}_{1}}{{r}^{5-1}} \\ & =1\cdot {{\left( \frac{1}{4} \right)}^{5-1}} \\ & =1\cdot {{\left( \frac{1}{4} \right)}^{4}} \\ & =\frac{1}{256} \end{align}\] Put \[n=6\]in\[{{a}_{n}}={{a}_{1}}{{r}^{n-1}}\]to find the sixth term of the given arithmetic sequence as follows: \[\begin{align} & {{a}_{6}}={{a}_{1}}{{r}^{6-1}} \\ & =1\cdot {{\left( \frac{1}{4} \right)}^{6-1}} \\ & =1\cdot {{\left( \frac{1}{4} \right)}^{5}} \\ & =\frac{1}{1024} \end{align}\] The given sequence is a geometric sequenceand the next two terms for the given sequence are\[\frac{1}{256}\ \text{and }\frac{1}{1024}\].