Question
Find six rational numbers between $2$ and $3$.
✓
Answer
$x = 2, y = 3$ and $n = 6$ $\text{d}=\frac{\text{y}-\text{x}}{\text{n}+1}=\frac{3-2}{6+1}=\frac{1}{7}$
Thus, the required numbers are $(\text{x}+\text{d}),(\text{x}+2\text{d}),(\text{x}+3\text{d}),...,(\text{x}+\text{n}\text{d})$
$=\Big(2+\frac{1}{7}\Big),\Big(2+2\times\frac{1}{7}\Big),\Big(2+3\times\frac{1}{7}\Big),$
$\Big(2+4\times\frac{1}{7}\Big),\Big(2+5\times\frac{1}{7}\Big),\Big(2+6\times\frac{1}{7}\Big)$
$=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$
Thus, the required numbers are $(\text{x}+\text{d}),(\text{x}+2\text{d}),(\text{x}+3\text{d}),...,(\text{x}+\text{n}\text{d})$
$=\Big(2+\frac{1}{7}\Big),\Big(2+2\times\frac{1}{7}\Big),\Big(2+3\times\frac{1}{7}\Big),$
$\Big(2+4\times\frac{1}{7}\Big),\Big(2+5\times\frac{1}{7}\Big),\Big(2+6\times\frac{1}{7}\Big)$
$=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$
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