Question
Give three rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$.
✓
Answer
Here $a =\frac{1}{3}, b=\frac{1}{2}, n =3$
$\therefore \frac{b-a}{n+1}=\frac{\frac{1}{2}-\frac{1}{3}}{3+1}=\frac{\frac{3}{6}}{4}=\frac{\frac{1}{6}}{4}=\frac{1}{24}$
$\therefore$ Three rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$ are
$\frac{1}{3}+\frac{1}{24}, \frac{1}{3}+2\left(\frac{1}{24}\right), \frac{1}{3}+3\left(\frac{1}{24}\right)$
$\frac{1}{3}+\frac{1}{24}, \frac{1}{3}+\frac{1}{12}, \frac{1}{3}+\frac{1}{8}$
$\frac{3}{8}, \frac{5}{12}, \frac{11}{24}$
$\therefore \frac{b-a}{n+1}=\frac{\frac{1}{2}-\frac{1}{3}}{3+1}=\frac{\frac{3}{6}}{4}=\frac{\frac{1}{6}}{4}=\frac{1}{24}$
$\therefore$ Three rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$ are
$\frac{1}{3}+\frac{1}{24}, \frac{1}{3}+2\left(\frac{1}{24}\right), \frac{1}{3}+3\left(\frac{1}{24}\right)$
$\frac{1}{3}+\frac{1}{24}, \frac{1}{3}+\frac{1}{12}, \frac{1}{3}+\frac{1}{8}$
$\frac{3}{8}, \frac{5}{12}, \frac{11}{24}$
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