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The terms in a geometric sequence are found by repeatedly multiplying the last term by a constant, called the common ratio.

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a)

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To find the common ratio, pick a term of the sequence and divide it by the previous term.

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We can calculate the common ratio using a table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $n$ $1$ $2$ $3$ $4$ $a_n$ $\\var{a*r}$ $\\var{a*r^2}$ $\\var{a*r^3}$ $\\var{a*r^4}$ Common ratio $\\displaystyle\\frac{\\var{a*r^2}}{(\\var{a*r})} = \\var{r}$ $\\displaystyle\\frac{\\var{a*r^3}}{\\var{a*r^2}} = \\var{r}$ $\\displaystyle\\frac{\\var{a*r^4}}{(\\var{a*r^3})} = \\var{r}$
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The common ratio is $\\var{d}$.

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b)

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The general formula for the $n^\\text{th}$ term of a geometric sequence is

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\$\\displaystyle {a_n=ar^{(n-1)}\\text{,}}\$

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where $a$ is the first term, and $r$ is the common ratio.

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So the formula for this sequence is

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\\\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} \

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c)

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We know from part b) that the $n^{th}$ term for this sequence is $a_n = \\var{a}(\\var{r})^n$.

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Therefore the $\\var{nth}^{th}$ term in the sequence is

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\\\begin{align} a_\\var{nth} &= \\var{a}(\\var{r})^\\var{nth}\\\\ &= \\var{a} \\times (\\var{{r}^{nth}})\\\\ &= \\var{{a}*{r}^{nth}}. \\end{align} \

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Find the common ratio for the following geometric series.

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$\\var{a*r}, \\var{a*r^2}, \\var{a*r^3}, \\var{a*r^4}\\ldots$

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Common ratio = [[0]]

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The formula for the $n^{th}$ term of a geometric sequence is

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\$\\displaystyle{ar^{(n-1)}}\$

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where $a$ is the first term in the sequence and $r$ is the common ratio.

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Find the formula for the  $n^{th}$ term in the sequence:

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$n^{th}$ term = [[0]]

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Find the value of the $\\var{nth}^{th}$ term.

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$a_\\var{nth}$ = [[0]]

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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.

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