General Form and nth Term of an AP

General form of an AP. Building on the first term $a$ and common difference $d$, every AP can be written as $a,\ a + d,\ a + 2d,\ a + 3d,\ a + 4d,\ \dots$ The key observation is that the number of $d$'s added is always one less than the position: position 1 has zero $d$'s, position 2 has one $d$, position 3 has two $d$'s, and so on.

The $n$th-term formula. Generalising that pattern gives the most-used result of the chapter: $a_n = a + (n - 1)d$. Here $a_n$ is the term in position $n$. This formula is a shortcut — it lets you find, say, the 50th term instantly without writing the first 49.

Why $(n - 1)$ and not $n$? Because no $d$ has been added yet at the very first term. Plug in $n = 1$: $a_1 = a + (1 - 1)d = a$, exactly the first term. The "$n - 1$" guarantees this. Writing $a + nd$ by mistake shifts every term by one place — the classic off-by-one slip in this topic.

Finding which term equals a given value. To check whether a number $N$ appears in an AP, set $N = a + (n - 1)d$ and solve for $n$. If $n$ turns out to be a positive integer, then $N$ is the $n$th term. If $n$ is a fraction, zero, or negative, $N$ is not a term of that AP. This is the standard "is 78 a term?" question.

Finding the last term and the number of terms. For a finite AP ending at $l$, the same formula reads $l = a + (n - 1)d$, which you can solve for $n$ to count the terms, or for $l$ if $n$ is known.

Finding $a$ and $d$ from two terms. If two terms are given (say $a_p$ and $a_q$), write each using the formula to get two equations and subtract: $a_q - a_p = (q - p)d$ gives $d$ directly. Then substitute back to find $a$, and you can reconstruct the whole AP.

The arithmetic-mean (middle-term) property. If three numbers $x, y, z$ are in AP, the middle one is the average of its neighbours: $y = \dfrac{x + z}{2}$, equivalently $2y = x + z$. This is the quickest way to fill in a missing middle term or to test three numbers for being in AP.

Terms counted from the end. A neat consequence: the $n$th term of a finite AP from the end is $l - (n - 1)d$. So the "3rd term from the end" of an AP ending at $l$ with common difference $d$ is $l - 2d$ — no need to know how many terms there are in total.

$a_n$ as a linear expression in $n$. Expanding the formula gives $a_n = dn + (a - d)$, which is a straight-line (linear) expression in $n$: the coefficient of $n$ is the common difference $d$, and the constant part is $a - d$. So if a sequence has $n$th term like $a_n = 3n + 1$, you can read off $d = 3$ at a glance, and recover $a$ from $a_1$. This also explains why plotting the terms of an AP against their position gives equally spaced points lying on a straight line.

Equally spaced terms stay in AP. Picking terms at regular gaps from an AP — say every 3rd term — again gives an AP, just with a larger common difference. This idea underlies many "the $p$th, $q$th, $r$th terms are in AP" style problems and is worth keeping in mind.

Common mistakes to avoid. (1) Using $a + nd$ instead of $a + (n - 1)d$. (2) Accepting a non-integer $n$ as a valid term position. (3) Sign errors when $d$ is negative — keep $(n - 1)d$ together and substitute carefully, e.g. $(n-1)(-3)$, not $-(n-1)3$ done loosely.