Respuesta :
Answer:
The numbers are 38, 39, and 40
Step-by-step explanation:
Let the 3 consecutive Integral numbers be;
x, x+1, x+2
We are informed that the sum of the 3 consecutive Integral numbers is 117. Therefore;
x + x+1 + x + 2 = 117
3x + 3 = 117
3x = 114
x = 38
x+1 = 39
x+2 = 40
Answer:
38, 39, 40
Step-by-step explanation:
Here's an interesting solution to this one. The formula to compute the nth triangular number, which is to say the sum of the first n consecutive integers, starting at 1 is
[tex]\frac{n(n+1)}{2}[/tex]
What if we wanted to start from a higher number, though? Say, 3. Well, we'd have to shift every number in the sequence up 2 (1, 2, 3 would become 3, 4, 5) so we'd be adding 2 n times. If we wanted to be more general, we could call that "shift amount" s, and our modified formula would now look like
[tex]\frac{n(n+1)}{2}+sn[/tex]
Now let's put this formula to the test. We know what our sum is here: it's 117. And we know what our n is too; we're finding 3 integers, so n = 3. This gives us the equation
[tex]\frac{3(3+1)}{2} +3s=117[/tex]
Solving this equation for s:
[tex]\frac{3(4)}{2} +3s=117\\\\\frac{12}{2}+3s=117\\ 6+3s=117\\3s=111\\s=37[/tex]
so our "shift amount" is 37, and our sequence gets shifted from 1, 2, 3 to 38, 39, 40.
But why?
This was a lot of setup for what seems like a disappointing payoff, but the real power with this approach is that we've actually just solved every problem of this type. Let's say you had to find the sum of 5 consecutive integers, and their sum was 70. Not a problem. Just set our n = 5 and solve:
[tex]\frac{5(6)}{2} +5s=70\\\\\frac{30}{2} +5s = 70\\15+5s=70\\5s=55\\s=11[/tex]
Which gives us a "shift" of 11 and the sequence 12, 13, 14, 15, 16 (which is exactly the sequence I came up with for this problem!)
