Components of an AP

An Arithmetic Progression is completely pinned down by just two numbers: its first term and its common difference. Once you know these two, every other term — and the whole sequence — is determined. The other quantities ($n$ and the last term) are then read off as needed.

First term ($a$). This is the starting value of the AP, the term in position 1. In $5, 9, 13, 17, \dots$ the first term is $a = 5$. It can be any number — positive, negative, a fraction or a decimal.

Common difference ($d$). The fixed amount added to move from one term to the next, found by subtracting any term from the term that follows it: $d = a_2 - a_1 = a_3 - a_2 = \dots$ Because it is "next minus previous", a falling sequence gives a negative $d$.

• $d > 0$: the AP is increasing (e.g. $3, 7, 11, 15$ with $d = 4$).
• $d < 0$: the AP is decreasing (e.g. $20, 17, 14, 11$ with $d = -3$).
• $d = 0$: every term is the same (a constant AP).

Number of terms ($n$). For a finite AP this counts how many terms there are; $n$ is always a positive whole number. The position of a term is its term-number: the first term sits at $n = 1$, the second at $n = 2$, and so on.

Last term ($l$, also written $a_n$). The final term of a finite AP. It follows from the others by $l = a + (n - 1)d$. The same formula, read backwards, lets you find $n$ when the first, last and common difference are known: $n = \dfrac{l - a}{d} + 1$.

How the terms are built. Each term stacks one more $d$ on top of $a$: $a_1 = a$, $a_2 = a + d$, $a_3 = a + 2d$, $a_4 = a + 3d$, and in general the $n$th term carries $(n - 1)$ copies of $d$. The pattern "position minus one, times $d$" is the heart of the next topic.

Finding $a$ and $d$ from two known terms. A frequent task: you are told, say, the 4th and 7th terms. Write each as $a + (\text{position} - 1)d$, getting two equations, then subtract to eliminate $a$ and solve for $d$; finally substitute back for $a$. This subtract-to-eliminate trick is used constantly in this chapter.

Reading the components off a real situation. In a word problem, the starting value of whatever is changing is $a$, and the fixed step-up or step-down is $d$. A salary of ₹12000 rising by ₹500 a year gives $a = 12000$, $d = 500$; a tank starting at 80 cm and losing 6 cm an hour gives $a = 80$, $d = -6$. Pinning down these two numbers first turns almost every AP word problem into a routine substitution.

The components are independent. Two different APs can share the same first term but differ in $d$, or share the same $d$ but start differently. Because of this, you genuinely need two pieces of information to fix an AP — typically two terms, or one term plus $d$, or the sum together with another fact. A single term on its own is never enough.

Common mistakes to avoid. (1) Computing $d$ as previous minus next, which reverses its sign. (2) Forgetting that the $n$th term carries $(n-1)$ — not $n$ — common differences, leading to an off-by-one error. (3) Treating a non-integer answer for $n$ as valid; the number of terms must be a positive integer, so a fractional $n$ means the value is not actually a term of that AP.