Question

7)

E) 10
The sum of any n consecutive integers is always equal to 100 less than the sum of the next n
consecutive integers. Find n.
A) 10
B) 8
C) 6
D) 13
E) 4

Answer 1

A) 10

Explanation:

Let the first integer be
`a+1`. Then we wish to solve for the following:

`(a+1) + (a+2) + \cdots + (a + n) + 100 = (a + (n + 1)) + (a + (n + 2)) + \cdots + (a + (2n))\\10a + (1 + 2 + \cdots + n) + 100 = 10a + ((n + 1) + (n + 2) + \cdots + (2n))\\(1)/(2)(n)(n + 1) + 100 = (1)/(2)(n)(3n + 1)\\n(n + 1) + 200 = n(3n + 1)\\n^2 + n + 200 = 3n^2 + n\\2n^2 = 200\\n^2 = 100\\n = 10 \quad \text{(because $n > 0$)}`