Thinking Mathematically (6th Edition)

Published by Pearson

Chapter 5 - Number Theory and the Real Number System - Chapter Summary, Review, and Test - Review Exercises - Page 336: 136

Answer

$\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{8},\,\,\frac{1}{16},\,\,\frac{1}{32}\ \text{and}\ \frac{1}{64}$.

Work Step by Step

For the second term put $n=2$ in the general formula stated above. \begin{align} & {{a}_{2}}={{a}_{1}}{{r}^{2-1}} \\ & =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{1}} \\ & =\frac{1}{2}\cdot \frac{1}{2} \\ & =\frac{1}{4} \end{align} For the third term put $n=3$ in the general formula stated above. \begin{align} & {{a}_{3}}={{a}_{1}}{{r}^{3-1}} \\ & =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{2}} \\ & =\frac{1}{2}\cdot \frac{1}{4} \\ & =\frac{1}{8} \end{align} For the fourth term put $n=4$ in the general formula stated above. \begin{align} & {{a}_{4}}={{a}_{1}}{{r}^{4-1}} \\ & =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{3}} \\ & =\frac{1}{2}\cdot \frac{1}{8} \\ & =\frac{1}{16} \end{align} For the fifth term put $n=5$ in the general formula stated above. \begin{align} & {{a}_{5}}={{a}_{1}}{{r}^{5-1}} \\ & =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{4}} \\ & =\frac{1}{2}\cdot \frac{1}{16} \\ & =\frac{1}{32} \end{align} For the sixth term put $n=6$ in the general formula stated above. \begin{align} & {{a}_{6}}={{a}_{1}}{{r}^{6-1}} \\ & =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{5}} \\ & =\frac{1}{2}\cdot \frac{1}{32} \\ & =\frac{1}{64} \end{align} The first six terms of the geometric sequence are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{8},\,\,\frac{1}{16},\,\,\frac{1}{32}\ \text{and}\ \frac{1}{64}$.

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