Skip to content
spmaddmath.com.my
Tuition

Study

SyllabusFormulasMethodsExam & PapersTools
LocationsPricingBlogOur TeachersContact
EN

Method · Progressions

Working with arithmetic progressions

An arithmetic progression adds a fixed common difference dd at every step. Use Tn=a+(n1)dT_n = a + (n-1)d for any term and Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a+(n-1)d] for the sum of the first nn terms.

What this method is for

An arithmetic progression, or AP, is a sequence in which every term is found by adding the same fixed number to the term before it. That fixed number is the common difference dd.

In Add Math you use this method to describe any situation that grows or shrinks in equal steps, a savings plan that adds the same amount each month, seats that increase by a fixed number in each row of a hall, or a stack whose height rises evenly.

Two workhorse formulas let you jump straight to an answer without listing every term. The first gives the value of any single term; the second gives the running total of the first nn terms.

Together they answer almost every AP question you will meet.

nth term
Tn=a+(n1)dT_n = a + (n-1)d
a = first term, d = common difference
sum of first n terms
Sn=n2[2a+(n1)d]S_n = \frac{n}{2}\left[2a + (n-1)d\right]

When to reach for it

Reach for this method whenever a list of numbers goes up or down by the same amount each time. Check by subtracting neighbouring terms: if T2T1=T3T2T_2 - T_1 = T_3 - T_2, the sequence is arithmetic and that shared value is dd.

Exam questions signal an AP with phrases such as 'increases by a constant amount', 'equal instalments', or a table of values with a steady gap.

If instead the terms are multiplied by a fixed ratio at each step, that is a geometric progression and needs different formulas. Watch closely for the word sum, it tells you to use SnS_n rather than TnT_n.

Deciding early which formula the question wants saves time and protects your working marks.

The steps

  1. 1

    Identify a and d

    Write down aa (the first term) and find dd by subtracting any term from the one after it: d=T2T1d = T_2 - T_1.

  2. 2

    Choose the right formula

    Use Tn=a+(n1)dT_n = a + (n-1)d if the question asks for a single term; use Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a+(n-1)d] if it asks for a sum.

  3. 3

    Set up equations if a or d is unknown

    If two terms are given, write each as a+(n1)da + (n-1)d, then solve the pair of simultaneous equations for aa and dd.

  4. 4

    Substitute carefully

    Put the known numbers in, keeping (n1)(n-1), not nn, inside the bracket.

  5. 5

    Simplify to a clean value

    Work through the arithmetic one line at a time and state the final answer with its units, if any.

  6. 6

    Check

    Verify by listing a few terms, or by testing your aa and dd in one of the terms the question gave you.

Worked example

Q1[5 marks]

In an arithmetic progression, the 3rd term is 11 and the 7th term is 23. Find the first term aa and the common difference dd, then find the sum of the first 12 terms.

Show worked solution

Write each given term using Tn=a+(n1)dT_n = a + (n-1)d.

T3=a+2d=11T_3 = a + 2d = 11
T7=a+6d=23T_7 = a + 6d = 23

Subtract the first equation from the second to remove aa:

(a+6d)(a+2d)=2311    4d=12    d=3(a+6d)-(a+2d) = 23-11 \;\Rightarrow\; 4d = 12 \;\Rightarrow\; d = 3

Substitute d=3d = 3 back into a+2d=11a + 2d = 11:

a+2(3)=11    a=5a + 2(3) = 11 \;\Rightarrow\; a = 5

Now find the sum of the first 12 terms with Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a+(n-1)d], taking n=12n = 12:

S12=122[2(5)+(121)(3)]=6[10+33]=6×43=258S_{12} = \frac{12}{2}\left[2(5) + (12-1)(3)\right] = 6\left[10 + 33\right] = 6 \times 43 = 258

So a=5a = 5, d=3d = 3, and the sum of the first 12 terms is 258258. As a quick check, the twelve terms are 5,8,11,,385, 8, 11, \ldots, 38; pairing first with last gives six pairs of 4343, and 6×43=2586 \times 43 = 258.

Common pitfalls

  • Using nn instead of (n1)(n-1) in the term formula, the first term needs zero steps of dd, so the multiplier is (n1)(n-1).
  • Mixing up TnT_n and SnS_n: a 'term' question wants a single value, a 'sum' question wants a total.
  • Getting the sign of dd wrong in a decreasing sequence, where dd is negative.
  • Solving the two-term equations by guessing instead of subtracting to eliminate aa cleanly.
  • Forgetting to substitute the found dd back to get aa, and stopping half way.

How a teacher helps

Most AP mistakes come down to one slip, a student writes nn where the formula needs (n1)(n-1), or reaches for TnT_n when the question quietly asks for a sum. In one-to-one lessons our teachers watch each line you write and stop at the exact moment the error appears, so the wrong habit never sets in.

We also train you to show the substitution line clearly, so your working is easy to follow and easy to credit line by line. Every teacher on spmaddmath.com.my is experienced.

Lessons are online and taught in English, matched to your own pace.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

What is the difference between an arithmetic and a geometric progression?

An arithmetic progression adds a fixed common difference dd at each step, while a geometric progression multiplies by a fixed common ratio rr. Test by subtracting terms for an AP and by dividing terms for a GP.

Which arithmetic progression formulas do I need to memorise?

The two key ones are Tn=a+(n1)dT_n = a + (n-1)d for the nnth term and Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a+(n-1)d] for the sum of the first nn terms. A second sum form, Sn=n2(a+l)S_n = \frac{n}{2}(a+l), is handy when the last term ll is known.

How do I find the common difference if I am only given two terms?

Write each term as a+(n1)da + (n-1)d, then subtract one equation from the other to eliminate aa. This leaves a single equation in dd; solve it, then substitute back to find aa.

How can I check my answer to an AP question?

List the first few terms using your aa and dd and confirm the gap is constant, or substitute your values back into one of the given terms. For a sum, a quick pairing check, first term plus last term, times the number of pairs, often confirms SnS_n.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Ready to get started?

Book a Trial Classfrom RM50/hr · One-hour paid trial · Same-day reply
Book a Trial ClassOne-hour paid trial · Same-day reply