Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.3 Geometric Sequences and Series - 14.3 Exercise Set - Page 911: 7

The series $4+20+100+500+2500+12,500$ is a geometric series.

A series having the same common ratio throughout the series is a geometric series. A series having the same common difference throughout the series is an arithmetic series. Consider the provided series, $4+20+100+500+2500+12,500$ The common ratio for the series is given by, $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Find the common ratio, $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Substitute $n=1$ in the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Thus, $\begin{align} & r=\frac{{{a}_{1+1}}}{{{a}_{1}}} \\ & =\frac{{{a}_{2}}}{{{a}_{1}}} \\ & =\frac{20}{4} \\ & =5 \end{align}$ Find the common ratio, $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Substitute $n=2$ in the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Thus, $\begin{align} & r=\frac{{{a}_{2+1}}}{{{a}_{2}}} \\ & =\frac{a3}{{{a}_{2}}} \\ & =\frac{100}{20} \\ & =5 \end{align}$ Find the common ratio, $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Substitute $n=3$ in the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Thus, $\begin{align} & r=\frac{{{a}_{3+1}}}{{{a}_{3}}} \\ & =\frac{{{a}_{4}}}{{{a}_{3}}} \\ & =\frac{500}{100} \\ & =5 \end{align}$ Find the common ratio, $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Substitute $n=4$ in the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Thus, $\begin{align} & r=\frac{{{a}_{4+1}}}{{{a}_{4}}} \\ & =\frac{{{a}_{5}}}{{{a}_{4}}} \\ & =\frac{2500}{500} \\ & =5 \end{align}$ Find the common ratio, $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Substitute $n=5$ in the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Thus, $\begin{align} & r=\frac{{{a}_{5+1}}}{{{a}_{5}}} \\ & =\frac{{{a}_{6}}}{{{a}_{5}}} \\ & =\frac{12500}{2500} \\ & =5 \end{align}$ So, this is a geometric series. Consider the provided series, $4+20+100+500+2500+12,500$ The common difference for the series is given by, ${{d}_{n}}={{a}_{n+1}}-{{a}_{n}}$ Find the common difference, ${{d}_{n}}={{a}_{n+1}}-{{a}_{n}}$ Substitute $n=1$ in the common ratio ${{d}_{n}}={{a}_{n+1}}-{{a}_{n}}$ Therefore, $\begin{align} & {{d}_{1}}={{a}_{1+1}}-{{a}_{1}} \\ & ={{a}_{2}}-{{a}_{1}} \\ & =20-4 \\ & =16 \end{align}$ Find the common difference, ${{d}_{n}}={{a}_{n+1}}-{{a}_{n}}$ Substitute $n=2$ in the common ratio ${{d}_{n}}={{a}_{n+1}}-{{a}_{n}}$ Thus, $\begin{align} & {{d}_{2}}={{a}_{2+1}}-{{a}_{1}} \\ & ={{a}_{3}}-{{a}_{1}} \\ & =100-20 \\ & =80 \end{align}$ Hence, ${{d}_{1}}\ne {{d}_{2}}$ So, this is not an arithmetic series. Thus, The series $4+20+100+500+2500+12,500$ is a geometric series.