Question

the sum of any n consecutive integers is always equal to 100 less than the sum of the next n consecutive integers. Find n.

Answer 1

When we have a given number k, the consecutive number to k will be (k + 1), the next consecutive will be (k + 2), and so on.

With this, we can find that the value of n is 100.

So here we have n consecutive numbers, let's assume we start with the number k, then the sum will be:

k + (k + 1) + ... + (k + n - 1)

Now we know that this is 100 less than the sum of the next n consecutive numbers.

For the next n consecutive numbers we need to start at k + 1, then we get:

(k + 1) + (k + 2) + ... + (k + n - 1) + (k + n)

If we take the difference we will get:

((k + 1) + (k + 2) + ... + (k + n - 1) + (k + n)) - (k + (k + 1) + ... + (k + n - 1)) = 100

As you can see, there are a lot of terms that are in both parentheses, then when we do that difference a lot of them cancel out.

So we will et:

((k + 1) + (k + 2) + ... + (k + n - 1) + (k + n)) - (k + (k + 1) + ... + (k + n - 1))

= (k + n) - k = k + n - k = n

And we knew that this difference must be equal to 100, then we have:

n = 100

We just found the value of n.

If you want to learn more, you can read:

brainly.com/question/1767889