Arithmetic Progression (AP)

A bus leaves Pune at 6 a.m. and arrives at every subsequent stop every $15$ minutes. The arrival times $6{:}00, 6{:}15, 6{:}30, 6{:}45, \dots$ form an arithmetic progression: each term differs from the previous by a fixed amount, $15$ minutes. This constant-difference pattern is the simplest and most ubiquitous kind of sequence.

Definitions

A sequence $a_1, a_2, a_3, \dots$ is an arithmetic progression (AP) if the difference $a_{n+1} - a_n$ is the same constant for every $n$. That constant is the common difference, denoted $d$.

If the first term is $a$, the AP looks like $a,\ a + d,\ a + 2d,\ a + 3d,\ \dots$

The $n$-th term

$\boxed{a_n = a + (n - 1) d.}$

This is a linear function of $n$ , APs are exactly the sequences whose general term is linear in $n$.

Sum of the first $n$ terms

Write the sum forward and backward: $S_n = a + (a + d) + (a + 2d) + \dots + (a + (n-1)d),$ $S_n = (a + (n-1)d) + (a + (n-2)d) + \dots + a.$ Add term by term: each of the $n$ pairs sums to $2a + (n-1)d$, so $2 S_n = n[2a + (n-1)d]$ and $\boxed{S_n = \frac{n}{2}[2a + (n-1)d].}$

If $\ell = a + (n-1)d$ is the last term, the formula also reads $S_n = \dfrac{n}{2}(a + \ell)$ , sum equals number of terms times the average of first and last.

Properties

Symmetry. Equidistant terms from the start and end sum to the same: $a_k + a_{n - k + 1} = a + \ell$ (for an AP of $n$ terms).

Reverse is also AP. If $a, a + d, \dots, \ell$ is an AP, so is $\ell, \ell - d, \dots, a$ , the common difference flips sign.

Mean property. In an AP, every term (except the first and last) is the arithmetic mean of its neighbours: $a_n = \dfrac{a_{n-1} + a_{n+1}}{2}$.

Three terms in AP. It is sometimes convenient to write three consecutive terms as $a - d, a, a + d$, four as $a - 3d, a - d, a + d, a + 3d$ , this makes sums simple and exploits symmetry.

Worked examples

Example 1. Find the $30$-th term of $5, 9, 13, 17, \dots$

Here $a = 5, d = 4$. So $a_{30} = 5 + 29 \cdot 4 = 121$.

Example 2. Find the sum of the first $50$ positive odd integers.

Odd integers: $1, 3, 5, \dots$ with $a = 1, d = 2$. So $a_{50} = 1 + 49 \cdot 2 = 99$ and $S_{50} = \dfrac{50}{2}(1 + 99) = 25 \cdot 100 = 2500$. (Equivalently, $50^2$.)

Example 3. Find three numbers in AP whose sum is $24$ and product is $440$.

Let the numbers be $a - d, a, a + d$. Sum $= 3a = 24 \Rightarrow a = 8$. Product $= a(a^2 - d^2) = 8(64 - d^2) = 440 \Rightarrow 64 - d^2 = 55 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3$. Numbers: $5, 8, 11$.

Example 4. In an AP, $a_5 = 30$ and $a_{12} = 65$. Find $a$ and $d$.

$a + 4d = 30$, $a + 11d = 65$. Subtracting: $7d = 35 \Rightarrow d = 5$, $a = 10$.

Example 5. Find the sum of all multiples of $7$ between $1$ and $1000$.

First: $7$, last: $994$. Number of terms: $\dfrac{994 - 7}{7} + 1 = 142$. Sum $= \dfrac{142}{2}(7 + 994) = 71 \cdot 1001 = 71{,}071$.

Try it yourself

1. Find the $20$-th term of $7, 13, 19, 25, \dots$
2. The $10$-th term of an AP is $32$, the $20$-th term is $72$. Find the AP.
3. Find the sum $4 + 9 + 14 + \dots + 254$.
4. How many terms of $3, 7, 11, \dots$ sum to $210$?
5. The sum of the first $n$ terms of an AP is $3n^2 + 5n$. Find the AP.
6. Find five numbers in AP whose sum is $25$ and whose product is $945$.
7. The $p$-th term of an AP is $q$ and the $q$-th term is $p$. Find the $(p + q)$-th term.
8. If $S_p = q$ and $S_q = p$ (with $p \ne q$), find $S_{p + q}$.
9. Find the sum of all two-digit multiples of $4$.
10. Find $\sum_{k=1}^{30} (5k - 3)$.
11. Three numbers are in AP. If $7$ is subtracted from the smallest, the numbers become a GP. Their sum is $36$. Find them.
12. Find the sum of $n$ terms of the AP $a, a + d, a + 2d, \dots$ when $a = -10, d = 4, n = 20$.

Pitfalls / Tricks

• Number of terms is $n = \dfrac{\ell - a}{d} + 1$. Don't forget the $+1$.
• For three terms in AP, use $a - d, a, a + d$. For four, use $a - 3d, a - d, a + d, a + 3d$.
• The sum formula $S_n = \frac{n}{2}(a + \ell)$ is the quickest , use it when both endpoints are known.
• Insight. Every AP is a linear function of the index. Knowing this, you can answer most questions just by setting up a linear equation.