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 points of the sequence plotted on a scatter graph where a strong positive correlation is formed

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 \(c\)?

A line of best fit is drawn and used to determine the 25th term of the sequence along with the equation of the graph
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 line of the original sequence and a translated version by 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 graph of the original sequence is drawn again on a larger scale and a line is drawn in order to determine the 195th term

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 \(y = \frac{97}{24}x-\frac{25}{24}\).

A second graph is drawn where the sequence starts at 3 and gets to 100 by the 25th term creating a much steeper gradient

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 \(100\)?