4.2 Q-4

Exercise 4.2 — Question 4

Problem

Find the arithmetic mean between the following pairs of numbers:

(i) $\sqrt{2}$ and $3\sqrt{2}$

(ii) $\frac{1}{3}$ and $\frac{1}{5}$

(iii) $a+b$ and $a-b$

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Key Formula

The arithmetic mean $A$ between two numbers $a$ and $b$ is:

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

Why it works: For $a$, $A$, $b$ to form an arithmetic sequence, the common difference must be constant: $A-a=b-A⟹2A=a+b⟹A=\frac{a+b}{2}$

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Solutions

Part (i): Between $\sqrt{2}$ and $3\sqrt{2}$

$A=\frac{\sqrt{2}+3\sqrt{2}}{2}=\frac{4\sqrt{2}}{2}=2\sqrt{2}$

Verification: $\sqrt{2},2\sqrt{2},3\sqrt{2}$ — common difference $=\sqrt{2}$ ✓

Part (ii): Between $\frac{1}{3}$ and $\frac{1}{5}$

$A=\frac{\frac{1}{3}+\frac{1}{5}}{2}=\frac{\frac{5+3}{15}}{2}=\frac{\frac{8}{15}}{2}=\frac{8}{30}=\frac{4}{15}$

Verification: $\frac{1}{3},\frac{4}{15},\frac{1}{5}$ — common difference $=\frac{4}{15}-\frac{5}{15}=-\frac{1}{15}$ ✓

Part (iii): Between $a+b$ and $a-b$

$A=\frac{(a+b)+(a-b)}{2}=\frac{2a}{2}=a$

Verification: $a+b,a,a-b$ — common difference $=-b$ ✓

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