a(n) = a + (n-1),
a(1)=a,
[a(1)]^2 = a^2 = 1*a(3) = a+2,
0 = a^2 - a - 2 = (a-2)(a+1),
a(1) = a = 2或a(1) = a = -1.
s(n) = na + n(n-1)/2,
a(1)a(9) = a(a+8) a^2 + 3a - 10 = (a+5)(a-2),
-5 < a = a(1) < 2.