We are told that Chiara's parents deposit a uniformly increasing amount of money into a savings account for Chiara every year on Chiara's birthday.
We are also given the amount of money that her parents deposit into the account on her first $3$ birthdays:
- On her $1^{st}$ birthday, they deposited $£\var{first}$ into the account.
- On her $2^{nd}$ birthday, they deposited $£\var{b[1]+first}$ into the account.
- On her $3^{rd}$ birthday, they deposited $£\var{b[1]*2+first}$ into the account.
a)
To calculate the amount of money Chiara's parents would deposit into the savings account on her 21st birthday, if her parents maintained this pattern, we use the equation
\[a_n=a_1+(n-1)d\text{,}\]
where
- $a_1$ is the first term;
- $a_n$ is the $n^{\text{th}}$ term;
- $d$ is the common difference between consecutive terms.
To identify the first term and common difference of the sequence we can use a table like the one below.
$n$ |
$1$ |
$2$ |
$3$ |
$a_n$ |
$\mathbf{\var{first}}$ |
$\var{b[1]+first}$ |
$\var{b[1]*2+first}$ |
First differences |
|
$\mathbf{\var{b[1]}}$ |
$\mathbf{\var{b[1]}}$ |
The first term and common difference have been highlighted in bold: $a_1 = \var{first}$ and $d = \var{b[1]}$.
Now we can use these to calculate $a_{21}$, giving us
\begin{align}
a_{21}&=\var{first}+\var{b[1]} \times (21-1) \\
&=\var{first+b[1]*(20)}\text{.} \\
\end{align}
So, assuming that Chiara's parents do maintain this pattern, on her 21st birthday her parents will deposit $£\var{first+b[1]*(20)}$ into the savings account.
b)
We are now asked to calculate the total amount of money that Chiara's parents will have added to this savings account over 21 years, including the money that her parents will deposit into the account on her 21st birthday.
This question involves calculating the sum using the equation
\[\sum\limits_{i=1}^n{a_i}=\frac{n}{2}(a_1+a_n)\text{.}\]
We know from part a) that
\begin{align}
a_1&=\var{first},\\
n&=21,\\
a_{21}&= \var{first+b[1]*(20)}.
\end{align}
Using our formula for the sum,
\begin{align}
\sum\limits_{i=1}^n{a_i}&=\frac{n}{2}(a_1+a_n)\\
&=\frac{\var{21}}{2}(\var{first}+\var{first+b[1]*(21-1)})\\
&=\var{21*(first+first+b[1]*(20))/2}\text{.}
\end{align}
Therefore, over 21 years Chiara's parents will have added a total of $£\var{21*(first+first+b[1]*(20))/2}$ to this savings account!