Arithmetic Mean and Selection of Terms

The Arithmetic Mean

If three numbers $a, b, c$ are in AP, then the middle one $b$ is the arithmetic mean (AM) of the other two:

$$b = \frac{a + c}{2}$$

This is just the ordinary average. The reason it works: in an AP, $b - a = c - b$, which rearranges to $2b = a + c$.

Example

The AM of 4 and 10 is $\dfrac{4 + 10}{2} = 7$. Indeed $4, 7, 10$ is an AP with $d = 3$.

Key Point: Three numbers are in AP if and only if the middle term equals the average of the outer two: $2b = a + c$.

[Board Important] To check if three numbers are in AP, verify $2 \times (\text{middle}) = (\text{first}) + (\text{last})$ — a one-line test.

Inserting Arithmetic Means

Sometimes we want to insert several numbers between two given numbers so the whole list becomes an AP. These inserted numbers are the arithmetic means.

Method

To insert $k$ arithmetic means between $a$ and $b$:

• The full AP has $k + 2$ terms: $a$, then $k$ means, then $b$.
• So $b$ is the $(k+2)$th term: $b = a + (k + 1)d$, giving $d = \dfrac{b - a}{k + 1}$.
• The means are $a + d, a + 2d, \dots, a + kd$.

Worked outline

Insert 3 means between 2 and 18: $d = \dfrac{18 - 2}{4} = 4$. Means: $6, 10, 14$. The AP is $2, 6, 10, 14, 18$.

Key Point: With $k$ means, the common difference is $d = \dfrac{b - a}{k + 1}$ (the gap is divided into $k+1$ equal steps).

[Board Important] Count carefully: inserting $k$ means creates $k + 1$ gaps between $a$ and $b$, so divide the difference by $k + 1$.

Choosing Terms in AP Cleverly

When a problem says 'three (or four) numbers are in AP' and gives conditions on their sum or product, choosing the terms symmetrically makes the algebra much easier.

For three terms in AP

Take them as $a - d, \ a, \ a + d$. Their sum is $3a$ — the $d$'s cancel!

For four terms in AP

Take them as $a - 3d, \ a - d, \ a + d, \ a + 3d$ (common difference $2d$). Their sum is $4a$.

Key Point: Symmetric choices ($a - d, a, a + d$ etc.) make the sum independent of $d$, so the sum condition gives $a$ immediately.

[Board Important] 'Three numbers in AP with sum 15' instantly gives $3a = 15 \Rightarrow a = 5$. Then a second condition (product, sum of squares) finds $d$.

Putting It Together

The symmetric-term trick turns a two-condition problem into two easy equations.

Worked outline

'Three numbers in AP have sum 15 and product 105. Find them.'

1. Let them be $a - d, a, a + d$. Sum $= 3a = 15 \Rightarrow a = 5$.
2. Product $= (5 - d)(5)(5 + d) = 105 \Rightarrow 5(25 - d^2) = 105 \Rightarrow 25 - d^2 = 21 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$.
3. Numbers: $3, 5, 7$ (or $7, 5, 3$).

Key Point: Find $a$ from the sum, then $d$ from the second condition.

[Board Important] Both $d = +2$ and $d = -2$ give the same set of numbers in different order — report the set.

Solved Examples

Example 1: Arithmetic mean

Find the arithmetic mean of 12 and 30.

Solution:

1. AM $= \dfrac{12 + 30}{2} = \dfrac{42}{2} = 21$.

Final Answer: 21.

Takeaway: AM is the average of the two numbers.

Example 2: Check three numbers in AP

Are 7, 11, 15 in AP?

Solution:

1. Check $2 \times 11 = 7 + 15$: $22 = 22$. ✓

Final Answer: Yes, they are in AP.

Takeaway: $2(\text{middle}) = (\text{first}) + (\text{last})$ is the AP test.

Example 3: Find the value for an AP

Find $x$ so that $x, 2x + 1, 7$ are in AP.

Solution:

1. $2(2x + 1) = x + 7 \Rightarrow 4x + 2 = x + 7$.
2. $3x = 5 \Rightarrow x = \dfrac{5}{3}$.

Final Answer: $x = \dfrac{5}{3}$.

Takeaway: Apply $2(\text{middle}) = (\text{first}) + (\text{last})$.

Example 4: Insert one mean

Insert one arithmetic mean between 8 and 20.

Solution:

1. One mean ⇒ it is just the AM: $\dfrac{8 + 20}{2} = 14$.

Final Answer: 14 (the AP is $8, 14, 20$).

Takeaway: A single inserted mean is the ordinary average.

Example 5: Insert several means

Insert 3 arithmetic means between 2 and 18.

Solution:

1. $d = \dfrac{18 - 2}{3 + 1} = \dfrac{16}{4} = 4$.
2. Means: $2 + 4 = 6$, $6 + 4 = 10$, $10 + 4 = 14$.

Final Answer: 6, 10, 14.

Takeaway: With $k$ means, $d = \dfrac{b - a}{k + 1}$.

Example 6: Three numbers, sum and product

Three numbers in AP have sum 15 and product 105. Find them.

Solution:

1. Let them be $a - d, a, a + d$. Sum $= 3a = 15 \Rightarrow a = 5$.
2. Product $= 5(25 - d^2) = 105 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$.
3. Numbers: $3, 5, 7$.

Final Answer: 3, 5, 7.

Takeaway: Symmetric terms make the sum give $a$ directly.

Example 7: Three numbers, sum of squares

Three numbers in AP have sum 12 and the sum of their squares is 56. Find them.

Solution:

1. $a - d, a, a + d$: sum $= 3a = 12 \Rightarrow a = 4$.
2. Sum of squares $= (4-d)^2 + 16 + (4+d)^2 = 48 + 2d^2 = 56 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$.
3. Numbers: $2, 4, 6$.

Final Answer: 2, 4, 6.

Takeaway: $(a-d)^2 + (a+d)^2 = 2a^2 + 2d^2$ simplifies neatly.

Example 8: Four numbers in AP

Four numbers in AP have sum 20 and the sum of their squares is 120. Find them.

Solution:

1. Take $a - 3d, a - d, a + d, a + 3d$. Sum $= 4a = 20 \Rightarrow a = 5$.
2. Sum of squares $= 2(a^2 + 9d^2) + 2(a^2 + d^2) = 4a^2 + 20d^2 = 120 \Rightarrow 100 + 20d^2 = 120 \Rightarrow d^2 = 1 \Rightarrow d = 1$.
3. Numbers: $2, 4, 6, 8$.

Final Answer: 2, 4, 6, 8.

Takeaway: For four terms, use $a \pm d, a \pm 3d$ (common difference $2d$).

Example 9: AM in a context

The arithmetic mean between two numbers is 25, and one number is 18. Find the other.

Solution:

1. AM $= \dfrac{18 + x}{2} = 25 \Rightarrow 18 + x = 50 \Rightarrow x = 32$.

Final Answer: 32.

Takeaway: Set the average equal to the AM and solve.

Example 10: Insert means — find a specific one

If 5 arithmetic means are inserted between 1 and 19, find the third mean.

Solution:

1. $d = \dfrac{19 - 1}{5 + 1} = \dfrac{18}{6} = 3$.
2. Third mean $= a + 3d = 1 + 9 = 10$.

Final Answer: 10.

Takeaway: The $k$th mean is $a + kd$, with $d = \dfrac{b-a}{(\text{means}) + 1}$.