Sn有最大值,说明d0.
a11/a10 < -1 => a11 < 0 且,a10 > 0,|a11| > |a10|.
a11 = a1 + 10d < 0,=>2a1 + 20d < 0
a10 = a1 + 9d > 0.=>2a1 + 18d > 0
=>a10 + a11 = 2a1 + 19d < 0
Sn = n*a1 + n(n-1)*d/2 < 0
=>a1 + (n-1)*d/2 < 0
=>2a1 + (n-1)*d