Introduction to arithmetic progressions

A sequence $a_1, a_2, a_3, \ldots$ is an arithmetic progression if there exists a fixed number $d$, the common difference, such that consecutive terms always differ by exactly $d$:

where: $a_1, a_2, a_3, \ldots$ are the terms of the list in order, and $d$ is the common difference — the same fixed number for the whole list. Equivalently, every term is built from the one before it by adding $d$:

where: $a_k$ is any term and $a_{k+1}$ is the term immediately after it; this recurrence is just the definition above written the other way round. The first term is usually called $a$ (or $a_1$). Three cases matter, based on the sign of $d$: an increasing AP has $d > 0$ (terms climb, like 10, 20, 30, …), a decreasing AP has $d < 0$ (terms fall, like 20, 16, 12, …), and a constant AP has $d = 0$ (every term equals $a$, like 5, 5, 5, …). Recognising which case you are in from the first two terms is often the fastest sanity check before doing any further work. To decide whether a given list of numbers is an AP, subtract each term from the one after it, all the way along the list, and compare the results. If every one of those differences comes out to the identical number, the list is an AP with that common difference. If even a single pair breaks the pattern, the whole list fails to be an AP — one bad gap is enough to disqualify it.

Starting only from the recurrence $a_{k+1} = a_k + d$ and a chosen first term $a$, show that every term of the progression can be written directly in terms of $a$, $d$, and its position in the list, without needing to work through every earlier term one by one.

Step 1. Write the first term as given: $a_1 = a$.

Step 2. Apply the recurrence once to get the second term: $a_2 = a_1 + d = a + d$.

Step 3. Apply the recurrence again, starting from $a_2$: $a_3 = a_2 + d = (a + d) + d = a + 2d$.

Step 4. Apply it once more, starting from $a_3$: $a_4 = a_3 + d = (a + 2d) + d = a + 3d$.

Step 5. A clear pattern has appeared: to reach $a_2$ you added $d$ once, to reach $a_3$ you added $d$ twice, to reach $a_4$ you added $d$ three times. In general, reaching the $k$th term from $a$ requires adding $d$ exactly $(k-1)$ times, so the whole progression can be listed out as

This confirms that once you know just the two numbers $a$ and $d$, you can write down as many terms of the AP as you like without repeating the addition step by hand each time — you simply count how many gaps of size $d$ separate the term you want from the first term.

Question. Write the first four terms of the AP whose first term is $a = 5$ and common difference is $d = -3$. Then check your answer using the consecutive-difference test.

Step 1. Use the pattern $a,\ a+d,\ a+2d,\ a+3d,\ldots$ derived above, with $a = 5$ and $d = -3$.

Step 2. Compute each term: $a_1 = a = 5$, $a_2 = a + d = 5 + (-3) = 2$, $a_3 = a + 2d = 5 + 2(-3) = 5 - 6 = -1$, and $a_4 = a + 3d = 5 + 3(-3) = 5 - 9 = -4$.

Step 3. Collect the results: the first four terms are $5,\ 2,\ -1,\ -4$.

Step 4. Check with the consecutive-difference test: $2 - 5 = -3$, $-1 - 2 = -3$, and $-4 - (-1) = -4+1= -3$. All three differences equal $-3$, matching the given $d$, so the answer is confirmed. The first four terms are $\mathbf{5,\ 2,\ -1,\ -4}$, and since $d = -3 < 0$, this is a decreasing AP.

Competitive stretch

Question. Two lists are given: (a) $2, 4, 8, 16, \ldots$ and (b) $7, 3, -1, -5, \ldots$. Decide which, if any, is an AP, and if it is, state $d$ and the next term.

Step 1. Test list (a): consecutive differences are $4-2=2$, $8-4=4$, $16-8=8$. These differences are $2, 4, 8$ — not the same number each time — so list (a) is not an AP. It is instead a doubling pattern, a geometric progression, which lies outside this chapter's scope.

Step 2. Test list (b): consecutive differences are $3-7=-4$, $-1-3=-4$, $-5-(-1)=-4$. All three differences equal $-4$, so list (b) is an AP with common difference $d = -4$.

Step 3. Find the next term of list (b) by adding $d$ to the last given term: $-5 + (-4) = -9$. List (b) is the AP, with $d = -4$ and next term $\mathbf{-9}$; list (a) is not an AP.

• Assuming any rising or falling list is automatically an AP. Only a list where every single consecutive gap is exactly the same fixed number qualifies — a doubling or squaring pattern can also look orderly but is not an AP.
• Stopping the check after only one pair. Confirming $a_2 - a_1 = d$ is not enough; a list can match for the first pair and then break the pattern later, so every consecutive pair must be checked.
• Losing track of sign when $d$ is negative. Writing $a + 3d$ with $d = -3$ means adding $-9$, not subtracting $9$ from the wrong step — keep the negative sign attached to $d$ throughout the calculation.
• Confusing "common difference" with "any difference." $d$ must specifically be $a_{k+1} - a_k$ in that order (later minus earlier); reversing the subtraction flips the sign and can lead to an incorrect classification of increasing versus decreasing.

Ask the tutor: If you want to test your own tricky-looking list of numbers or double-check whether it forms an AP, ask the AI tutor to work through it with you step by step.