4.2 Q-14

Exercise 4.2 — Question 14

Problem

Find the arithmetic mean between $\sqrt{2}$ and $3\sqrt{2}$.

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Solution

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

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

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

Here, $a=\sqrt{2}$ and $b=3\sqrt{2}$:

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

Verification: The sequence $\sqrt{2},2\sqrt{2},3\sqrt{2}$ has a constant common difference of $\sqrt{2}$, confirming it is arithmetic. ✓

$\boxed{A=2\sqrt{2}}$