The difference of two natural numbers is 3 and the difference of their reciprocals is

Solution:

Let the required natural numbers be x and (x + 3).

Now, x < x + 3

$\therefore \frac{1}{x}>\frac{1}{x+3}$

According to the given condition,

$\frac{1}{x}-\frac{1}{x+3}=\frac{3}{28}$

$\Rightarrow \frac{x+3-x}{x(x+3)}=\frac{3}{28}$

$\Rightarrow \frac{3}{x^{2}+3 x}=\frac{3}{28}$

$\Rightarrow x^{2}+3 x=28$

$\Rightarrow x^{2}+3 x-28=0$

$\Rightarrow x^{2}+7 x-4 x-28=0$

$\Rightarrow x(x+7)-4(x+7)=0$

$\Rightarrow(x+7)(x-4)=0$

$\Rightarrow x+7=0$ or $x-4=0$

$\Rightarrow x=-7$ or $x=4$

∴ x = 4             (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.