Question

Find four consecutive intergers such that the product of the 1st interger and the 3rd interger is 25 greater than the product of -13 and the fourth integer
Answers: -14,-13,-12,-11

Answer 1

Let’s say that your sequence of four numbers starts at some number n.
Then the four numbers are n, n+1, n+2, and n+3.

You’re asked to find that number n
so that n(n+2) = 25 + (-13)(n+3).

One way to solve this is to notice that this is a quadratic equation in n.
You’re guaranteed that an integer solution exists,
so you’d expect to be able to factor the quadratic expression.

If you try, you’ll simplify it down to n^2 + 15n + 14 = 0.
And this is (n + 14)(n + 1) = 0.

So n can be either -14 or -1.

If n = -14, then you get the sequence -14, -13, -12, -11.
If you let n = -1, then you get the sequence -1, 0, 1, 2.

You can check that both sequences satisfy our condition on the numbers.