Arithmetic mean and special tricks

A few small but powerful ideas often unlock AP problems in $30$ seconds. This last topic collects them.

Arithmetic mean

The arithmetic mean (AM) of two numbers $a$ and $b$ is $\text{AM} = \frac{a + b}{2}.$

Notably, $a, \text{AM}, b$ are three consecutive terms of an AP , the AM sits exactly in the middle. So the AM property is just the middle-term test for an AP.

More generally, if we want $n$ numbers between $a$ and $b$ in AP (so that $a, A_1, A_2, \ldots, A_n, b$ is an AP of $n + 2$ terms), the common difference is $d = \frac{b - a}{n + 1}.$

This is how to insert $n$ arithmetic means between two given numbers.

Symmetric notation for AP terms

A clever way to set up problems involving three or four unknowns in AP:

Three terms in AP: $a - d, \quad a, \quad a + d$ . The middle term is the average. The sum is $3 a$ .

Four terms in AP with common difference $2 d$ : $a - 3 d, \quad a - d, \quad a + d, \quad a + 3 d$ . The sum is $4 a$ .

Five terms in AP: $a - 2 d, \quad a - d, \quad a, \quad a + d, \quad a + 2 d$ . Sum is $5 a$ .

Why is this useful? Because in many problems the sum of the unknowns is given. With symmetric notation, the sum is just a multiple of $a$ , and you instantly know $a$ . The constraint involving $d$ usually then gives one more equation.

Connecting tricks

Trick 1: If $m$ th term is $n$ and $n$ th term is $m$ , then the $(m + n)$ th term is $0$ . Quick proof: $a + (m - 1) d = n$ and $a + (n - 1) d = m$ . Subtract: $(m - n) d = n - m \Rightarrow d = -1$ . Then $a + (m - 1)(-1) = n \Rightarrow a = n + m - 1$ . So $a + (m + n - 1) d = n + m - 1 - (m + n - 1) = 0$ .

Trick 2: If $S_m = S_n$ for $m \ne n$ , then $S_{m + n} = 0$ . A quick algebraic check using $S_n = \dfrac{n}{2}[2 a + (n - 1) d]$ proves this.

Trick 3: The $k$ th term equals the average of the $(k - p)$ th and $(k + p)$ th terms. That is, $a_k = \dfrac{a_{k - p} + a_{k + p}}{2}$ . Useful in "middle term" problems.

Worked examples

Example 1. Find three numbers in AP whose sum is $24$ and whose product is $440$ .

Let them be $a - d, a, a + d$ . Sum $= 3 a = 24 \Rightarrow a = 8$ .

Product: $a(a^2 - d^2) = 440 \Rightarrow 8(64 - d^2) = 440 \Rightarrow 64 - d^2 = 55 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3$ .

Numbers: $5, 8, 11$ (or $11, 8, 5$ , same set).

Example 2. Find four numbers in AP whose sum is $20$ and the sum of whose squares is $120$ .

Let them be $a - 3 d, a - d, a + d, a + 3 d$ . Sum $= 4 a = 20 \Rightarrow a = 5$ .

Sum of squares: $(a - 3 d)^2 + (a - d)^2 + (a + d)^2 + (a + 3 d)^2 = 4 a^2 + 20 d^2 = 120$ .

So $100 + 20 d^2 = 120 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1$ .

Numbers: $2, 4, 6, 8$ .

Example 3. Insert four arithmetic means between $3$ and $23$ .

We need $3, A_1, A_2, A_3, A_4, 23$ to be in AP , total $6$ terms. So $d = (23 - 3)/5 = 4$ . Means: $7, 11, 15, 19$ .

Example 4. If the $m$ th term of an AP is $n$ and the $n$ th term is $m$ , find the $(m + n)$ th term.

Using Trick 1: $a_{m + n} = 0$ .

Example 5. If $S_5 = S_{15}$ for an AP, find $S_{20}$ .

Using Trick 2: $S_{5 + 15} = S_{20} = 0$ .

Try it yourself

Insert five arithmetic means between $8$ and $26$ .
Find three numbers in AP whose sum is $15$ and product is $80$ .
Four numbers in AP have sum $32$ and sum of extremes $24$ . Find them.
If the $10$ th term of an AP is $52$ and the $17$ th term is $87$ , find the $25$ th term.
The sum of three numbers in AP is $-3$ and their product is $8$ . Find them.
The angles of a triangle are in AP; the smallest is $30^\circ$ . Find the others.
Find the AM of $-7$ and $13$ . Then write the AP that has them as extreme terms with $3$ means between.
The angles of a quadrilateral are in AP whose common difference is $10^\circ$ . Find them.
If $S_m = S_n$ for an AP (with $m \ne n$ ), prove that $S_{m+n} = 0$ .
If the $p$ th, $q$ th, and $r$ th terms of an AP are $a, b, c$ respectively, prove that $a(q - r) + b(r - p) + c(p - q) = 0$ .

Pitfalls / Insight

Symmetric notation only works when the sum of the unknowns is known.
AM lies between $a$ and $b$ in value , if your computed mean is outside, you've made an error.
Number of "means" between $a$ and $b$ is the count of inserted terms, not the total terms.

Insight. APs are about linear growth. Whenever a problem has linear growth and a sum, an AP formula likely solves it in one or two lines. Practice spotting the structure , that is the chapter's whole exam value.