The sum of two numbers is 9.

Let one numbers be x then other (9 − x).

Then according to question

$\frac{1}{x}+\frac{1}{(9-x)}=\frac{1}{2}$

$\Rightarrow \frac{(9-x)+x}{x(9-x)}=\frac{1}{2}$

$\Rightarrow \frac{9}{x(9-x)}=\frac{1}{2}$

By cross multiplication

$\Rightarrow 18=x(9-x)$

$\Rightarrow x^{2}-9 x+18=0$

$\Rightarrow x^{2}-6 x-3 x+18=0$

$\Rightarrow(x-6) x-3(x-6)=0$

$\Rightarrow(x-6)(x-3)=0$

$\Rightarrow x=6,3$

Since, x being a number,

Therefore,

When x = 6 then

(9 − x) = (9 − 6) = 3

When x = 3 then

(9 − x) = (9 − 3) = 6

Thus, two consecutive number be either 3, 6 or 6, 3.