the consecutive numbers are 1/x and 1/(x+1)
you can graph that function and then graph y = 17/12 and you will find that the intersections of both lines are not integers.
the function you would graph would be y = 1/x + 1/(x+1).
here's the graph.
it shows that the intersection is not an integer.
to show you that this method works, we'll pick one where the reciprocal of the integers do add up to what the result is.
for example: 1/5 + 1/6 = 6/30 + 5/30 = 11/30
your equation would be y = 1/x + 1/(x+1) again, only this time your intersection would be 11/30 rather than 17/12.
here's the graph.
the graph shows that the intersection is at x = 5.
this agrees with the setup because the first of the integers is x = 5.
therefore the sum of 1/5 + 1/6 = 11/30 as calculated beforehand and as shown on the graph.
bottom line is that your problem has no solution.
there are no positive integers such that 1/x + 1/(x+1) = 17/12.
in fact, assuming that the smallest integers is 1, there is only one set of integers where the result would be greater than 1.
that would be 1/1 + 1/2 = 3/2 = 1.5.
any other integers would have a result smaller than 1.
1/2 + 1/3 = 5/6
1/3 + 1/4 = 7/12
etc.
you can figure this out for yourself by solving the equation y = 1/x + 1/(x+1)
just pick different values of x and you will see that the result is less than 1 for all values of x > 1.
here's that graph.
the graph is accurate.
for example, when x = 5, you get 1/5 + 1/6 = 11/30 = .367 as shown on the graph.