For an AP with negative common difference (a decreasing AP with a positive first term), the sum $S_n$ is maximum when:
$n$ is as large as possible
$a_n$ is the last non-negative term
$n=1$
$d>0$ instead
Step-by-Step Solution
Key Concept: Adding a negative term decreases the sum, so the sum is maximised right before terms turn negative.
As long as terms are non-negative, adding them increases $S_n$; once terms turn negative, adding them decreases $S_n$. So $S_n$ is maximum at the last non-negative term. [1.0 Mark]
Correct Answer: $a_n$ is the last non-negative term