Arithmetic Progressions: Counting the Terms Without Losing One

The question

The integers between 150 and 450 that are multiples of 8 form an arithmetic progression.

(a) Find the number of terms in this progression. [2 marks]

(b) Find the sum of all the integers between 150 and 450 that are multiples of 8. [2 marks]

The common wrong answer

Most students find the first and last terms correctly. The first multiple of 8 above 150 is 152, and the last multiple of 8 below 450 is 448. Then they count the terms like this:

n = \frac{448 - 152}{8} = \frac{296}{8} = 37

and use 37 in the sum formula:

S_{37} = \frac{37}{2}(152 + 448) = \frac{37}{2} \times 600 = 11\,100

The mistake is in the count. $\frac{448 - 152}{8}$ counts the gaps between the terms, not the terms themselves. There is always one more term than there are gaps, so 37 is one short, and the sum comes out 300 too small. Examiners see this kind of slip often: the Pearson Edexcel 4PM1 examiner report for November 2020 noted many candidates who used the right method for an arithmetic series sum but the wrong number of terms.

The correct working

(a) The first term is $a = 152$ , the common difference is $d = 8$ and the last term is 448. Use the formula for the $n$ th term:

T_n = a + (n - 1)d

Set the $n$ th term equal to the last term:

\begin{aligned} 152 + (n - 1)(8) &= 448 \\ (n - 1)(8) &= 296 \\ n - 1 &= 37 \\ n &= 38 \end{aligned}

There are 38 terms.

(b) Now use the sum formula with $n = 38$ . Since we know the first and last terms, the quickest form is:

S_n = \frac{n}{2}(a + l)

S_{38} = \frac{38}{2}(152 + 448) = 19 \times 600 = 11\,400

So the sum is 11 400.

You can check this with the other form of the formula, $S_n = \frac{n}{2}[2a + (n - 1)d]$ :

S_{38} = \frac{38}{2}[2(152) + 37(8)] = 19 \times 600 = 11\,400

You may see the $n$ th term written as $T_n$ or as $u_n$ (as in the Cambridge List of formulas). They mean the same thing.

Why it works

Think of a fence with posts every 8 metres. A 24-metre fence has $\frac{24}{8} = 3$ gaps but 4 posts, because there is a post at both ends. The terms of a progression are the posts, and the common difference is the gap.

Take the small progression 8, 16, 24, 32. The gap from first to last is $32 - 8 = 24$ , and $\frac{24}{8} = 3$ , but there are 4 terms. That is exactly why the formula has $(n - 1)$ in it: to get from the first term to the $n$ th term, you add $d$ only $n - 1$ times.

So “last minus first, divided by $d$ ” gives $n - 1$ , and you must add 1 to get $n$ .

How to spot it next time

Always find $n$ by solving $a + (n - 1)d = \text{last term}$ , rather than dividing the gap by $d$ in your head.
Test your method on a tiny progression you can count by hand, such as 8, 16, 24, 32 (4 terms).
Check the words “between”: the first and last terms are the first and last multiples inside the range (152 and 448 here), not 150 and 450 themselves.