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Find the common ratio of a given geometric sequence, write down the formula for the nth term and use it to calculate a given term in the sequence.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "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$\\var{a}$ | \n$\\var{a*r}$ | \n$\\var{a*r^2}$ | \n$\\var{a*r^3}$ | \n
$a_n \\div a_{n-1}$ | \n\n | $\\var{r}$ | \n$\\var{r}$ | \n$\\var{r}$ | \n
The common ratio is $\\var{r}$.
\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\\[ a_n = \\simplify[]{ {a}*{r}^(n-1) } \\text{.} \\]
\nWe know from part b) that the formula for the $n^\\text{th}$ term is $a_n = \\simplify[]{ {a}*{r}^{(n-1)}}$.
\nTherefore the $\\var{n}^\\text{th}$ term in the sequence is
\n\\begin{align}
a_\\var{n} &= \\var{a} \\times \\var{r}^{\\var{n-1}} \\\\
&= \\var{a*r^(n-1)}
\\end{align}
The first term
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\nThe range is picked so that the number is between 1,000 and 1,000,000.
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\n$\\var{a}, \\var{a*r}, \\var{a*r^2}, \\var{a*r^3}, \\ldots$
\nCommon ratio: [[0]]
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\nWrite down the formula for the $n^\\text{th}$ term in the sequence
\n$a_n = $ [[0]]
", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "The formula for the $n^\\text{th}$ term of a geometric sequence is
\n\\[ a_n = ar^{(n-1)} \\]
\nwhere $a$ is the first term in the sequence and $r$ is the common ratio.
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\n$a_\\var{n} =$ [[0]]
", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "a*r^(n-1)", "maxValue": "a*r^(n-1)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Geometric Sequence - negative ratio ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bradley Bush", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1521/"}], "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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