Given a geometric sequence, find the common ratio (negative in this question), write down the formula for the nth term and use it to calculate a given term.
"}, "preamble": {"css": "", "js": ""}, "advice": "The terms in a geometric sequence are found by repeatedly multiplying the last term by a constant, called the common ratio.
\nTo find the common ratio, pick a term of the sequence and divide it by the previous term.
\nWe can calculate the common ratio using a table:
\n| $n$ | \n$1$ | \n$2$ | \n$3$ | \n$4$ | \n
| $a_n$ | \n$\\var{a*r}$ | \n$\\var{a*r^2}$ | \n$\\var{a*r^3}$ | \n$\\var{a*r^4}$ | \n
| Common ratio | \n\n | $\\displaystyle\\frac{\\var{a*r^2}}{(\\var{a*r})} = \\var{r}$ | \n$\\displaystyle\\frac{\\var{a*r^3}}{\\var{a*r^2}} = \\var{r}$ | \n$\\displaystyle\\frac{\\var{a*r^4}}{(\\var{a*r^3})} = \\var{r}$ | \n
The common ratio is $\\var{d}$.
\nThe general formula for the $n^\\text{th}$ term of a geometric sequence is
\n\\[\\displaystyle {a_n=ar^{(n-1)}\\text{,}}\\]
\nwhere $a$ is the first term, and $r$ is the common ratio.
\nSo the formula for this sequence is
\n\\[
\\begin{align}
a_n&=ar^{(n-1)}\\\\
&=\\var{a*r}\\times(\\var{r})^{(n-1)}\\\\
&=(\\var{a} \\times (\\var{r}))(\\var{r})^{n-1}\\\\
&=\\var{a}(\\var{r})^n\\text{.}
\\end{align}
\\]
We know from part b) that the $n^{th}$ term for this sequence is $a_n = \\var{a}(\\var{r})^n$.
\nTherefore the $\\var{nth}^{th}$ term in the sequence is
\n\\[
\\begin{align}
a_\\var{nth} &= \\var{a}(\\var{r})^\\var{nth}\\\\
&= \\var{a} \\times (\\var{{r}^{nth}})\\\\
&= \\var{{a}*{r}^{nth}}.
\\end{align}
\\]
Find the common ratio for the following geometric series.
\n$\\var{a*r}, \\var{a*r^2}, \\var{a*r^3}, \\var{a*r^4}\\ldots$
\nCommon ratio = [[0]]
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The formula for the $n^{th}$ term of a geometric sequence is
\n\\[\\displaystyle{ar^{(n-1)}}\\]
\nwhere $a$ is the first term in the sequence and $r$ is the common ratio.
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\n$n^{th}$ term = [[0]]
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\n$a_\\var{nth}$ = [[0]]
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