What is the least number of terms of the AP 4, 11, 18, 25, ... whose sum exceeds 400?

$a=4,d=7$, so $S_n=\frac{n}{2}$$[8+7(n-1)]=\frac{7n^2+n}{2}$. Then $S_{10}=355$, which is still below 400, while $S_{11}=429$, which is above it, so 11 terms are needed. Stopping at 10 fails the "exceeds" test, and 12 and 14 overshoot without being the least such value. Note that the question asks for a count of terms, not for the sum 429 itself.