The formula for the $n^\text{th}$ term, $a_n$, of an arithmetic sequence is

\[ a_n=a_1+(n-1)d \text{.} \]

$a_1$ is the first term, and $d$ is the common difference between adjacent terms.

a)

In the given sequence, the common difference is $\var{a1+d} - \var{a1} = \var{d}$, and the first term is $\var{a1}$.

So, the formula for this sequence is

\[ a_n = \var{a1} + (n-1) \times \var{d} \text{.} \]

b)

\[ a_\var{small} = \var{a1} + (\var{small}-1) \times \var{d} = \var{a1+(small-1)*d} \text{.} \]

c)

\[ a_\var{large} = \var{a1} + (\var{large}-1) \times \var{d} = \var{a1+(large-1)*d} \text{.} \]

In an arithmetic sequence, the difference between two adjecent terms is always the same.

a)

The difference between the first two terms in this sequence is $\var{m[0]*(index[0]+1)} - \var{m[0]*(index[0])} = \var{m[0]}$.

So to get from one term to the next, add $\var{m[0]}$.

The complete sequence is 

\[ \var{m[0]*(index[0])}, \; \var{m[0]*(index[0]+1)}, \; \var{m[0]*(index[0]+2)}, \; \underline{\var{m[0]*(index[0]+3)}}, \; \var{m[0]*(index[0]+4)}, \; \underline{\var{m[0]*(index[0]+5)}}, \; \var{m[0]*(index[0]+6)} \]

b)

The difference between the first two terms in this sequence is $\var{m[1]*(index[1]-1)} - \var{m[1]*(index[1])} = \var{-m[1]}$.

So to get from one term to the next, subtract $\var{m[1]}$.

The complete sequence is 

\[ \var{m[1]*(index[1])}, \; \var{m[1]*(index[1]-1)}, \; \underline{\var{m[1]*(index[1]-2)}}, \; \var{m[1]*(index[1]-3)}, \; \var{m[1]*(index[1]-4)}, \; \underline{\var{m[1]*(index[1]-5)}}, \; \var{m[1]*(index[1]-6)} \]

In an arithmetic sequence, the difference between two adjecent terms is always the same.

Call the common difference $d$, and the first term in the sequence $a_0$. Then the sequence goes as follows:

\[ a_0, \; a_0+d, \; a_0+2d, \; a_0+3d, \; \ldots \]

The difference between a term in the sequence and the term $n$ places along is $n \times d$.

a)

The difference between the first two terms is $\var{n[0]*(index[0]+1)} - \var{n[0]*index[0]} = \var{n[0]}$.

So the common difference is $\var{n[0]}$.

b)

We're not given any adjacent terms of this sequence, but we are given some terms two palces apart.

\begin{align}
2d &= \var{n[1]*(index[1]+2)} - \var{n[1]*index[1]} \\
&= \var{2*n[1]} \\
d &= \var{n[1]}
\end{align}

The common difference is $\var{n[1]}$.

c)

Again we're not given any adjacent terms of this sequence, but we have two terms three places apart.

\begin{align}
3d &= \var{n[2]*(index[2]+3)} - \var{n[2]*index[2]} \\
&= \var{3*n[2]} \\
d &= \var{n[2]}
\end{align}

The common difference is $\var{n[2]}$.

We are told that Brooke's parents deposit a uniformly increasing amount of money into a savings account for Brooke every year on Brooke'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 Brooke's parents would deposit into the savings account on her 21^st 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.

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 Brooke's parents do maintain this pattern, on her 21^st 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 Brooke'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 21^st 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 Brooke's parents will have added a total of $£\var{21*(first+first+b[1]*(20))/2}$ to this savings account!

We know that every $\var{d}^{\text{th}}$ ticket after the first receives strawberry ice cream. So, the sequence of ticket numbers which get strawberry ice cream starts like this:

\[ 1, \var{d+1}, \var{2d+1}, \var{3d+1}, \ldots \]

The numbers on the tickets for strawberry ice cream form an arithmetic sequence: the first term is $1$ and the common difference is $\var{d}$.

We can write down a formula for the $n^{\text{th}}$ term in this sequence and rearrange it to find how many customers received strawberry ice cream before Ruby.

The general formula for an arithmetic sequence is

\[a_n = a_1 + (n-1)d \]

where

• $a_n$ is the $n^{th}$ term;
• $a_1$ is the first term;
• $d$ is the common difference between consecutive terms.

We know that $a_n = \var{1+d*index}$, $a_1 = 1$, and $d = \var{d}$, and we want to find $n$.

Substituting these values from the formula gives

\begin{align}
a_n &= a_1 + (n-1)d \\
\var{1+d*index} &= 1 + (n -1)\var{d}\,.
\end{align}

Now we rearrange this to find $n$:

\begin{align}
\var{1+d*index-1} &= \var{d}n - \var{d} \\
\var{d*index +d} &= \var{d}n \\
n &= \var{(d*index + d)/d}.
\end{align}

This is the number of people who received strawberry ice cream up to and including Ruby. Removing Ruby leaves $\var{index}$ customers who received strawberry ice cream before Ruby did.

Both of these sequences are linear, or arithmetic, sequences. To find formulas for these sequences we need to identify their first terms and common differences.

a)

The formula for the $n^\text{th}$ term of an arithmetic sequence is

\[ a_n = a_1 + (n-1)d \text{.} \]

$a_1$ is the first term and $d$ the common difference between consecutive terms, which we need to identify.

We can find these by drawing up a table of $a_n$ against $n$, and the differences between consecutive terms.

The first term and common difference have been highlighted in bold; we can use these to write the formula for the sequence.

\begin{align}
a_n &= a_1+(n-1)d \\
&= \var{m[1]*2}+(n-1)\times\var{m[1]} \\
&= \var{m[1]}n + \var{m[1]}\text{.}
\end{align}

b)

Similar to the part a), we can identify $a_1$ and $d$ for this sequence by drawing a table of $a_n$ against $n$.

The first term and common difference have been highlighted in bold; we can use these to form the formula for the sequence. 

\begin{align}
a_n &=a_1+(n-1)d \\
&=\var{m[2]*8+2}+(n-1)\times\var{-m[2]} \\
&=-\var{m[2]}n + \var{m[2]*9+2}\text{.}
\end{align}

We are given the arithmetic sequence

\[\var{m*first_index}, \var{m*(first_index+1)}, \var{m*(first_index+1)}, \ldots, \var{last_term} \text{.} \]

a)

Let $n$ be the number of terms in the sequence. Then $a_n = \var{last_term}$.

To find $n$, we must rearrange the formula for the $n^\text{th}$ term

\[a_n=a_1+(n-1)d\text{.}\]

The first term is $a_1 = \var{first_term}$ and the common difference is $d = \var{first_term+m} - \var{first_term} = \var{m}$.

So we have

\begin{align}
d&=\var{m} \text{,} \\
a_1&= \var{first_term} \text{,} \\
a_n&=\var{last_term} \text{.}
\end{align}

We now substitute these values into the formula for the $n^\text{th}$ term and rearrange to find $n$.

\begin{align}
\var{last_term}&=\var{first_term}+\var{m}(n-1) \\
\var{last_term}-\var{first_term}&=\var{m}(n-1) \\
\frac{\var{last_term-first_term}}{\var{m}}&=(n-1) \\
n&=\var{(last_term-first_term)/m}+1 = \var{num_terms}\text{.}
\end{align}

b)

The partial sum of the first $n$ terms of an arithmetic sequence is given by

\[\sum\limits_{i=1}^n{a_i}=\frac{n}{2}(a_1+a_n)\text{.}\]

We know from part a) that

\begin{align}
n&= \var{num_terms} \text{,} \\
a_1&= \var{first_term} \text{,} \\
a_n&= \var{last_term} \text{.}
\end{align}

We substitute these values into the formula, obtaining

\begin{align}
\sum\limits_{i=1}^n{a_i}&=\frac{n}{2}(a_1+a_n)\\
&= \frac{\var{num_terms}}{2}(\var{first_term}+\var{last_term})\\
&= \simplify[]{ ({num_terms}*({first_term+last_term}))/2} \\
&= \var{partial_sum} \text{.}
\end{align}