Question

If the sum of the reciprocals of two consecutive positive numbers is $\frac{21}{110}$ then find the numbers.

Answer 1

Let the two consecutive positive numbers be $x$ and $x+1$.

Then according to the problem,

$\frac{1}{x}+\frac{1}{x+1}=\frac{21}{110}$

or, $\frac{2x+1}{x(x+1)}=\frac{21}{110}$

or, $220x+110=21x^2+21x$

or, $21x^2-199x-110=0$

or, $21x^2-210x+11x-110=0$

or, $21x(x-10)+11(x-10)=0$

or, $(x-10)(21x+11)=0$

or, $x=10$. [ Since $x$ is positive]

So the numbers are $10,11$.