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Part a)

To find the $i^{th}$ term for a specific $i$ (in this case $\\var{tr}$) we need to recall the following formulae:

$\\displaystyle\\sum\\limits_{i=1}^nar^{i-1}$ $_{..(I)}$

$a_i=ar^{i-1}$ $_{..(II)}$

$a$ is the first term in the sequence. Namely, $\\var{t1}$.

To find the common ratio, $r$, simply divide one term by its predecessor.

The known values for $a$, $r$ and $i$ are then subsituted into $_{(II)}$.

Part b)

To 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 $i$.

In this case,

$a_i=\\var{t1}(\\simplify{-1/{td}})^{i-1}$

Part c)

To find the sum of the first $n$ (in this case $\\var{tsum}$) terms of a geometric series, we turn to the following formula:

$\\displaystyle\\sum\\limits_{i=1}^nar^{i-1}=a\\left(\\frac{1-r^n}{1-r}\\right)$ $_{..(III)}$

The sum is found by subsituting the known values for $a$, $r$ and $n$ into $_{(III)}$.

Part d)

The sum to infinity of a geometric series can be found by first taking the usual sum from $_{(III)}$ and setting $n$ to $\\infty$:

$\\displaystyle\\sum\\limits_{i=1}^{\\infty}ar^{i-1}=a\\left(\\frac{1-r^\\infty}{1-r}\\right)$ $_{..(III)}$

When $\\left|r\\right|<0$, note that $r^\\infty$ tends to $0$. The above formula simplifies:

$a\\left(\\frac{1-0}{1-r}\\right)=a\\left(\\frac{1}{1-r}\\right)=\\left(\\frac{a}{1-r}\\right)$

$\\therefore\\;\\;\\;\\displaystyle\\sum\\limits_{i=1}^{\\infty}ar^{i-1}=\\left(\\frac{a}{1-r}\\right)$ $_{..(IV)}$

The final sum is calculated by substituting the values for $a$ and $r$ into $_{(IV)}$.

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Find the $\\var{tr}$^thterm. Give your answer accurate to 3 significant figures.

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Find the $i^{th}$ term

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Find the sum of the first $\\var{tsum}$ terms. Give your answer accurate to 2 decimal places.

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Find the sum to infinity of the series. Give your answer accurate to 2 decimal places.

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For the geometric series:

$\\var{t1} - \\simplify{{t1}/{td}} + \\simplify{{t1}/{td}^2} - \\simplify{{t1}/{td}^3} +...$

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