Question 1035962
as far as i can tell, there is no integer that satisfies this problem.


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.


<img src = "http://theo.x10hosting.com/2016/052502.jpg" alt="$$$" </>


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.


<img src = "http://theo.x10hosting.com/2016/052503.jpg" alt="$$$" </>


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.


<img src = "http://theo.x10hosting.com/2016/052504.jpg" alt="$$$" </>


the graph is accurate.


for example, when x = 5, you get 1/5 + 1/6 = 11/30 = .367 as shown on the graph.