Problem: The sum of n terms of an arithmetic progression is 153, and the common difference is 2. If the first term is an integer, and n>1, then the number of possible values for n is:
Answer Choices:
A. 2
B. 3
C. 4
D. 5
E. 6
Solution:
Let a be the first term of the progression. Then 2n​[a+(a+2(n−1))]=153, or n2+n(a−1) −153=0. Pairs of factors of -153 are −1,153;1,−153,−9,17;9,−17;−3,51;3,−51. Therefore, the possible values of a −1 are +152,−152,+8,−8,+48,−48 and the possible values of n are 1,153,9,17,3, and 51. The value n=1 is rejected, sn that there are five permissible values.