Rich example

## Solution

Find the last term of the sequence that starts with $3$, has a common difference of $\frac{1}{2}$ and has $25$ terms.

• In what ways can this problem be solved?

Let’s write the sequence numerically: $3, 3\frac{1}{2}, 4, 4\frac{1}{2},...$

The first term is $3$, and after that there will be $24$ other terms. $\frac{1}{2}$ is being added for each term, so over $24$ terms, $12$ will be added. Therefore the $25$th term will be $3 + 12 = 15$.

How does this example help us explain why the nth term of an arithmetic sequence is $a_n = a_1 + (n-1)d$? Why is it $n-1$ rather than $n$?

We could also plot the points of the sequence on a graph.

The sequence is a linear relationship, so we can draw a linear graph through the points with gradient of $\frac{1}{2}$. The equation of the line is $y = \frac{1}{2}x + c$.

What is the value of $c$?

The equation of the line is $y = \frac{1}{2}x + 2\frac{1}{2}$, and so the $25$th term will be, \begin{align*} y &= \frac{1}{2} \times 25 + 2\frac{1}{2} \\ &= 15. \end{align*}

How does thinking about the sequence graphically and the equation $y = \frac{1}{2}x + 2\frac{1}{2}$, link to $a_n = a_1 + (n-1)d$, the formula for the nth term of an arithmetic sequence?

Change only one aspect of the sequence in order to make the last term $100$.

• What aspects can be changed?
• What would they change to?

We are told about three aspects of the sequence: the first term, the number of terms, and the common difference.

#### The first term

If we change the first term, then we still have a common difference of $\frac{1}{2}$ and $25$ terms, so difference between the first and last term will still be $12$. Therefore, to make the last term $100$ the first term would be $88$.

If we think about the old and new sequences as linear graphs then we have a translation of $85$ units in the $y$ direction.

#### The number of terms

If we change the number of terms then we have the same intitial sequence, with a first term of $3$ and a common difference of $\frac{1}{2}$. This means that we can look at the original graph of $y = \frac{1}{2}x + 2\frac{1}{2}$, and we want to know when $y = 100$. Solving for $x$ gives us $x = 195$, i.e. there would be $195$ terms in this sequence.

#### The common difference

If we change the common difference we want a sequence that starts at $3$ and gets to $100$ by the $25$th term. This means we are adding on $97$ in 24 terms. Therefore the common difference will be $\frac{97}{24}$. Graphically we have a line that still passes through the point $(1,3)$, but also passes through $(25, 100)$ giving the equation $y = \frac{97}{24}x-\frac{25}{24}$.

We might have tried to change the common difference in a different way. We could make it a common ratio, and change the sequence from arithmetic to geometric. Why can’t a geometric sequence with a first term of $3$, and common ratio of $\frac{1}{2}$ ever have a nth term of $100$?

Is it possible to have other sequences that have a first term of $3$, and a $25$th term of $100$?