The sum of k consecutive integers is 41. If the least integer is -40, then k =
A.) 40
B.) 41
C.) 80
D.) 81
E.) 82
A.) 40
B.) 41
C.) 80
D.) 81
E.) 82
(-40) + (-39) + (-38) +...+ (-1) + 0 + 1 +...+ 38 + 39 + 40 = 0.infiniti007 wrote:The sum of k consecutive integers is 41. If the least integer is -40, then k =
A.) 40
B.) 41
C.) 80
D.) 81
E.) 82
All of the values in red cancel out, yielding a sum of 0.
To yield a sum of 41, one more integer is needed:
(-40) + (-39) + (-38) + ...+ (-1) + 0 + 1...+ 38 + 39 + 40 + 41 = 41.
In the sum above, there are 40 integers to the left of 0, 41 integers to the right of 0, and 0 itself, for a total of k=82 integers.
The correct answer is E.
