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Part a)
\n$a$ is the first term in the sequence. Namely, $\\var{t1}$
\nTo find the common ratio, $r$, simply divide one term by its predecessor.
\nPart b)
\nTo find the $i^{th}$ term, $a_i$, for a specific $i$, recall the sigma notation for the sum of a geometric series:
\n$\\displaystyle\\sum\\limits_{i=1}^nar^{i-1}$ $_{..(I)}$ also, note that any individual term is found by: $a_i=ar^{i-1}$ $_{..(II)}$
\nTo calculate $a_\\var{tr}$, substitute the known values of $a$, $r$ and $i$ into $_{(II)}$ and simplify.
\nPart c)
\nWhen using the sigma notation $\\sum$ as in $_{(I)}$, each individual term is often labelled with the letter $i$ and the total number of terms in the series is defined as the value for $n$. It is important to recognise what each letter is referring to in each instance to avoid confusion.
\nTo find the general $i^{th}$ term, $a_i$, formula of a geometric series, we substitute known values of $a$ and $r$ into $_{(II)}$ to produce the formula in $i$.
\nIn this case,
\n$a_i=\\var{t1}(\\var{r})^{i-1}$
\nPart d)
\nTo find the sum of the first $n$ terms of a geometric series, we turn to the following formula:
\n$\\displaystyle\\sum\\limits_{i=1}^nar^{i-1}=a\\left(\\frac{1-r^n}{1-r}\\right)$ $_{..(III)}$
\nThe sum is found by subsituting the known values for $a$, $r$ and $n$ into $_{(III)}$.
\n", "rulesets": {}, "parts": [{"prompt": "State the $a$ and $r$
\n$a$=[[1]]
\n$r$=[[0]]
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", "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "{t1}(1-({r})^{tsum})/((1-{r}))", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "extensions": [], "statement": "For the geometric series:
\n$\\var{t1}$+$\\simplify{{t1}*{r}}$+$\\simplify{{t1}*{{r}^2}}$+$\\simplify{{t1}*{{r}^3}}$+...
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