Understanding arithmetic and geometric progressions

A progression, you may also hear "sequence", is simply a list of numbers built by a fixed rule. Each number is a term, the first term is written $a$, and the term in position $n$ is written $T_{n}$.

The whole chapter rests on one question: what is the rule that gets you from one term to the next? Add Math cares about exactly two answers.

• If you get to the next term by adding a fixed number, it is an arithmetic progression, and that fixed number is the common difference $d$.
• If you get to the next term by multiplying by a fixed number, it is a geometric progression, and that fixed number is the common ratio $r$.

The very first thing to do in any question is decide which kind you are looking at. Check the gap between terms: if $T_{2}-T_{1}$ equals $T_{3}-T_{2}$, it is arithmetic; if the ratio $T_{2}/T_{1}$ equals $T_{3}/T_{2}$, it is geometric.

Everything after that follows from this single choice.

In an arithmetic progression, you start at $a$ and add $d$ again and again. So the terms are $a$, $a+d$, $a+2d$, $a+3d$, and so on.

Notice the pattern in the multiples of $d$: the second term has one $d$, the third has two. That is where the $n-1$ in the formula comes from, the $n$th term has added $d$ a total of $n-1$ times.

$$T_{n} = a + (n-1)d$$

To add up the first $n$ terms, there is a neat formula. It is worth reading it as "the number of terms, times the average of the first and last term", which is exactly what it says once you look closely.

$$S_{n} = \frac{n}{2}\left[\,2a + (n-1)d\,\right]$$

In a geometric progression, you start at $a$ and multiply by $r$ again and again. The terms are $a$, $ar$, $ar^{2}$, $ar^{3}$, and so on.

Again count the powers: the second term has $r^{1}$, the third has $r^{2}$, so the $n$th term has $r^{n-1}$.

$$T_{n} = ar^{\,n-1}$$

The sum of the first $n$ terms has two equivalent forms; use whichever keeps the numbers positive and tidy. Both are valid for $r \neq 1$.

$$S_{n} = \frac{a\left(r^{\,n} - 1\right)}{r - 1} = \frac{a\left(1 - r^{\,n}\right)}{1 - r}$$

Here is the part that feels almost magical the first time: if the common ratio $r$ is a fraction between $-1$ and $1$, you can add up infinitely many terms of a geometric progression and still get a finite answer. Think of $8 + 4 + 2 + 1 + \tfrac{1}{2} + \dots$, each term is half the last, and the running total creeps towards 16 without ever passing it.

$$S_{\infty} = \frac{a}{1 - r}, \qquad \text{valid only when } |r| < 1$$

Most lost marks in this chapter come from a handful of predictable slips. Knowing them beforehand protects more marks than extra practice on the parts you already do well.

• Reaching for the wrong family, using an arithmetic formula on a geometric progression, or the reverse. Always confirm whether terms are added or multiplied first.
• The $n-1$ trap: the $n$th term uses $n-1$, not $n$. $T_{5}$ of a GP is $ar^{4}$, not $ar^{5}$.
• Using $S_{\infty}$ when $|r| \geq 1$, where no sum to infinity exists.
• Sign errors when $d$ or $r$ is negative, keep the brackets and substitute carefully.

Progressions is a chapter where the whole thing clicks once the add-versus-multiply distinction is truly solid, and stays muddled while the formulas float around unattached to that idea. Working one-to-one, a teacher can watch how your child decides which family a question belongs to, catch the exact place the $n-1$ or the ratio check goes wrong, and rebuild the reasoning so the formulas feel earned.

Our teachers are experienced; lessons are online and taught in English, while SPM papers are set bilingually in Bahasa Melayu and English.

If progressions are a sticking point, you are welcome to begin with a one-hour paid trial class at the teacher's own rate (from RM50 per hour, depending on experience). We won't promise a grade, but we can help turn this from a chapter of half-remembered formulas into one of the more dependable sources of marks on the paper.