**Concept: The question tests on the ability to solve Sequence problems**

**Solution:**

Sum of an AP = n/2∗[2a+(n−1)d]n2

Converting all values into months

Sum = 250 * 12 = 3000, a = 7 * 12 = 84, d = 3

Substituting, we get 3000 = n/2∗[2∗84+(n−1)3]

6000 = n[168 + 3n - 3]

3n^2+165n−6000=0

n^2+55n−2000=0

n2+80n−25n−2000=0

n(n + 80) - 25(n + 80) = 0

(n + 80)(n - 25) = 0

n = 25 or - 80

Option E