Arithmetic progressions

Some sequences of numbers are wild and unpredictable; others obey a simple rule of "add the same amount every time." An arithmetic progression (AP) is the second kind — a list like $7, 11, 15, 19, \dots$ where each term is the previous one plus a fixed constant. That single idea, repeated, lets you jump straight to the 100th term of a list without writing out ninety-nine terms first, and lets you add up thousands of terms in one line of algebra instead of a thousand additions. This article builds the two formulas that do all the work — the general term $a_n$ and the sum $S_n$ — shows where the sum formula actually comes from, and works through the same style of problem ICSE papers ask: find a specific term, find a sum, and (the twist examiners love) work backwards from a given sum to find how many terms were added. Along the way you will see why AP problems sometimes turn into quadratic equations, and how to avoid the handful of slips that cost the most marks.

The core insight

An arithmetic progression is a sequence with a constant common difference $d$ between consecutive terms, so its $n$ th term is $a_n = a + (n-1)d$ and the sum of its first $n$ terms is $S_n = \tfrac{n}{2}[2a + (n-1)d]$ .

The core idea

An arithmetic progression is a sequence in which every term after the first is obtained by adding a fixed number, the common difference $d$ , to the term before it. If the first term is $a$ , the AP looks like

a,\ a+d,\ a+2d,\ a+3d,\ \dots

Finding $d$ . Since consecutive terms differ by exactly $d$ , it is always computed as (any term) minus (the term just before it):

d = a_2 - a_1 = a_3 - a_2 = \dots

where: $a_1, a_2, a_3, \dots$ are the 1st, 2nd, 3rd, ... terms of the sequence. Checking only one pair of terms is risky — a sequence can coincidentally match for one gap and fail the next — so a genuine AP needs every consecutive gap to be equal.

The general (nth) term. To reach the $n$ th term from the first term, you add $d$ a total of $(n-1)$ times, not $n$ times, because the first term itself already counts as "zero steps" of $d$ :

a_n = a + (n-1)d

where: $a_n$ is the $n$ th term, $a$ is the first term, $d$ is the common difference, and $n$ is a positive whole number (the position in the list).

The sum of the first $n$ terms. Adding up all the terms from $a$ to $a_n$ gives:

S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]

where: $S_n$ is the sum of the first $n$ terms. If the last (nth) term $l = a_n$ is already known, the same sum is often quicker to write as $S_n = \dfrac{n}{2}(a+l)$ — useful whenever the question hands you the last term directly instead of $d$ . The figure below shows the picture behind all of this: equally spaced terms on a number line, each one exactly $d$ apart from its neighbours, which is the entire content of "arithmetic progression" in one glance.

Figure

A number line showing the terms 7 11 15 19 of an arithmetic progression with equal gaps of 4 marked between consecutive terms

The sum formula, from first principles

Derivation AWhy the sum formula pairs first and last terms

The goal is to derive $S_n = \dfrac{n}{2}[2a+(n-1)d]$ from nothing but the definition of an AP, using a pairing trick that needs no calculus and works for any first term, common difference, or number of terms — the same trick Gauss is said to have used as a schoolboy to add $1$ to $100$ in seconds.

Step 1. Write the sum $S_n$ forwards, from the first term to the last term $l = a_n$ , and then write the very same sum backwards, from the last term to the first:

S_n = a + (a+d) + (a+2d) + \dots + l

S_n = l + (l-d) + (l-2d) + \dots + a

Both lines equal $S_n$ because they list the identical set of terms, just in opposite order.

Step 2. Add the two lines term by term, position by position. In every position, one line contributes a term that has grown by some multiple of $d$ above $a$ , and the other line contributes the matching term that has shrunk by exactly the same multiple of $d$ below $l$ — so the $+d$ and $-d$ pieces cancel in every single pair, leaving $(a+l)$ from each of the $n$ positions:

2S_n = (a+l) + (a+l) + \dots + (a+l) \quad (n \text{ times}) = n(a+l)

Step 3. Divide both sides by 2 to isolate $S_n$ :

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

This is already a usable formula whenever the last term $l$ is known directly.

Step 4. Replace $l$ using the general-term formula $l = a_n = a+(n-1)d$ derived above, to get a version that needs only $a$ , $d$ and $n$ — no separate calculation of the last term required:

S_n = \dfrac{n}{2}\big[a + a+(n-1)d\big] = \dfrac{n}{2}\big[2a+(n-1)d\big]

Check: for $a=1$ , $d=1$ , $n=100$ (Gauss's problem), $S_{100} = \tfrac{100}{2}[2(1)+99(1)] = 50\times101 = 5050$ , matching the well-known answer for $1+2+\dots+100$ .

Worked example

Question: Find the 15th term and the sum of the first 15 terms of the AP $7, 11, 15, 19, \dots$

Step 1. Identify $a$ and $d$ . Here $a = 7$ and $d = 11-7 = 4$ .

Step 2. Find the 15th term using $a_n = a+(n-1)d$ :

a_{15} = 7 + (15-1)(4) = 7+56 = 63

Step 3. Find the sum of the first 15 terms using $S_n = \dfrac{n}{2}[2a+(n-1)d]$ :

S_{15} = \dfrac{15}{2}\big[2(7)+(15-1)(4)\big] = \dfrac{15}{2}\big[14+56\big] = \dfrac{15}{2}(70) = 525

Step 4. Cross-check with the alternate form $S_n = \dfrac{n}{2}(a+l)$ : $S_{15} = \dfrac{15}{2}(7+63) = \dfrac{15}{2}(70) = 525$ — matches. So $a_{15} = 63$ and $S_{15} = 525$ .

Competitive stretch

How many terms of the AP $5, 8, 11, 14, \dots$ must be taken so that their sum is $440$ ?

Step 1. Here $a=5$ , $d=3$ , and $S_n=440$ is given while $n$ is unknown, so substitute into $S_n = \dfrac{n}{2}[2a+(n-1)d]$ :

440 = \dfrac{n}{2}\big[10+3(n-1)\big] \implies 880 = n(3n+7) \implies 3n^2+7n-880=0

Step 2. Solve this quadratic in $n$ with the quadratic formula ( $a=3,b=7,c=-880$ ): $n = \dfrac{-7\pm\sqrt{49+10560}}{6} = \dfrac{-7\pm103}{6}$ , giving $n=16$ or $n=-\tfrac{110}{6}$ . The negative root is rejected since a number of terms cannot be negative. So $n=16$ terms are needed, confirmed by $S_{16} = \dfrac{16}{2}(5+50)=440$ .

Common traps

Mixing up $a_n$ (a single term) with $S_n$ (the running sum) — the question always tells you which one it wants, so underline it first.
Using $nd$ instead of $(n-1)d$ in the general term — the first term takes zero steps of $d$ , so the offset is always one less than $n$ .
Forgetting to reject a negative or non-integer root when a word problem turns into a quadratic in $n$ , since $n$ must be a positive whole number.
Computing $d$ inconsistently when terms are fractional, negative, or given out of order — always take (a later term) minus (the term right before it), never the reverse.
Treating a real-world pattern (savings that grow by a fixed amount each month, seats that increase by a fixed number in each row) as an AP without first checking that the increase really is constant — some "increasing" patterns are actually geometric or irregular, and only a constant additive step qualifies.

Ask the tutor: want to practise turning a word problem into an AP, or check your own sum calculation step by step — ask the AI tutor to walk through your specific sequence.

Why is the common difference sometimes negative?

An AP is only required to have a constant difference between consecutive terms — nothing says that difference must be positive. If each term is smaller than the one before, like $20, 17, 14, 11, \dots$ , then $d = 17-20 = -3$ , and the formulas $a_n = a+(n-1)d$ and $S_n = \tfrac{n}{2}[2a+(n-1)d]$ still work exactly as written, simply producing decreasing terms and, eventually, negative ones.

Do I always have to use the messy formula with 2a plus (n-1)d, or is there a shortcut?

Whenever the last term $l$ of the sum is already known or easy to find, use the shorter form $S_n = \tfrac{n}{2}(a+l)$ instead — it needs one less calculation. Use the longer $S_n = \tfrac{n}{2}[2a+(n-1)d]$ when you only know $a$ , $d$ and $n$ , and the last term has not been worked out yet.

How do I know a sequence is really an arithmetic progression and not something else, like a geometric progression?

Compute the difference between every consecutive pair of terms, not just the first pair. If that difference is the same constant value all the way through, it is an AP; a geometric progression instead has a constant ratio (not difference) between consecutive terms, so checking a subtraction versus a division is the quickest way to tell the two apart.