// Numbas version: finer_feedback_settings {"name": "Patricia's copy of future value - ordinary annuity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Patricia's copy of future value - ordinary annuity", "tags": [], "metadata": {"description": "Financial maths. Future value of an ordinary annuity.", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "
Suppose you are given $\\$\\var{C}$ to deposit into a savings account at the end of each {period[2]} for $\\var{years}$ years. If the interest rate is $\\var{ipa}\\%$ per annum compounding {period[0]}, how much money is in the savings account after $\\var{years}$ years?
", "advice": "You are asked to find the future value of an ordinary annuity (since the payments are at the end of each period). Therefore we will use the future value of an ordinary annuity formula
\n$\\displaystyle A=R(\\frac{(1+i)^n-1}{i})$
\nwhere $A$ is the future value, $R$ is the amount deposited per period, $i$ is the interest rate per period, and $n$ is the number of periods.
\nIn our situation we have,
\n$R=\\var{C}$,
\n$i=\\frac{\\var{ipa}\\%}{\\var{period[1]}}=\\frac{\\var{ipadec}}{\\var{period[1]}}$, $i=\\var{ipa}\\%=\\var{ipadec}$,
\n$n=\\var{years}\\times \\var{period[1]}=\\var{n}$, $n=\\var{n}$,
\nand therefore we have
\n$\\displaystyle A=\\var{C}(\\frac{(1+\\frac{\\var{ipadec}}{\\var{period[1]}})^\\var{n}-1}{\\frac{\\var{ipadec}}{\\var{period[1]}}})$
\nCalculating this we find
\n$\\begin{align}A&\\approx \\var{F}\\\\&=\\$\\var{Frounded}\\quad \\text{(to the nearest cent)}\\end{align}$
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\n\n$\\$$ [[0]] (to the nearest cent)
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\n$\\displaystyle A=R\\left[\\frac{(1+i)^n-1}{i}\\right]$
\nwhere $A$ is the future value, $R$ is the amount deposited per period, $i$ is the interest rate per period, and $n$ is the number of periods.
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