Answer:
A) 10
Step-by-step explanation:
Let the first integer be [tex]a+1[/tex]. Then we wish to solve for the following:
[tex](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))\\\frac{1}{2}(n)(n + 1) + 100 = \frac{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 $n > 0$)}[/tex]
