The difference of two natural numbers is 5 and the difference of their reciprocals is

Solution:

Let the required natural numbers be x and (x + 5).

Now, x < x + 5

$\therefore \frac{1}{x}>\frac{1}{x+5}$

According to the given condition,

$\frac{1}{x}-\frac{1}{x+5}=\frac{5}{14}$

$\Rightarrow \frac{x+5-x}{x(x+5)}=\frac{5}{14}$

$\Rightarrow \frac{5}{x^{2}+5 x}=\frac{5}{14}$

$\Rightarrow x^{2}+5 x=14$

$\Rightarrow x^{2}+5 x-14=0$

$\Rightarrow x^{2}+7 x-2 x-14=0$

$\Rightarrow x(x+7)-2(x+7)=0$

$\Rightarrow(x+7)(x-2)=0$

$\Rightarrow x+7=0$ or $x-2=0$

$\Rightarrow x=-7$ or $x=2$

∴ x = 2             (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.