Notes: ``8th term:` 1/2, -1/8, 1/32, -1/128...` Find the indicated term of the geometric sequence.

Friday, December 5, 2014

You may use the following formula, such that:

$\displaystyle{a}_{{n}}={a}_{{1}}\cdot{{r}}^{{{n}-{1}}}$

Replacing 8 for n, $\displaystyle\frac{{1}}{{2}}$ for $\displaystyle{a}_{{1}}$ yields:

$\displaystyle{a}_{{8}}=\frac{{1}}{{2}}\cdot{{\left({r}\right)}}^{{7}}$

You may find the ratio r, such that:

$\displaystyle{r}=\frac{{{a}_{{2}}}}{{{a}_{{1}}}}\Rightarrow{r}=\frac{{-\frac{{1}}{{8}}}}{{\frac{{1}}{{2}}}}\Rightarrow{r}=-\frac{{1}}{{4}}$

$\displaystyle{a}_{{8}}=\frac{{1}}{{2}}\cdot{{\left(-\frac{{1}}{{4}}\right)}}^{{7}}\Rightarrow{a}_{{8}}=-\frac{{1}}{{{{2}}^{{15}}}}$

Hence, the indicated term $\displaystyle{a}_{{8}}$ is $\displaystyle{a}_{{8}}=-\frac{{1}}{{{{2}}^{{15}}}}$

You may use the following formula, such that:

$\displaystyle{a}_{{n}}={a}_{{1}}\cdot{{r}}^{{{n}-{1}}}$

Replacing 8 for n, $\displaystyle\frac{{1}}{{2}}$ for $\displaystyle{a}_{{1}}$ yields:

$\displaystyle{a}_{{8}}=\frac{{1}}{{2}}\cdot{{\left({r}\right)}}^{{7}}$

You may find the ratio r, such that:

$\displaystyle{r}=\frac{{{a}_{{2}}}}{{{a}_{{1}}}}\Rightarrow{r}=\frac{{-\frac{{1}}{{8}}}}{{\frac{{1}}{{2}}}}\Rightarrow{r}=-\frac{{1}}{{4}}$

$\displaystyle{a}_{{8}}=\frac{{1}}{{2}}\cdot{{\left(-\frac{{1}}{{4}}\right)}}^{{7}}\Rightarrow{a}_{{8}}=-\frac{{1}}{{{{2}}^{{15}}}}$

Hence, the indicated term $\displaystyle{a}_{{8}}$ is $\displaystyle{a}_{{8}}=-\frac{{1}}{{{{2}}^{{15}}}}$

Notes