Question

1：等差数列前n项和是sn,若-am＜a1＜-am+1…（求讲解第一题和第二题）

Answer 1

题1,

a(n) = a+ (n-1)d.

s(n) = na + n(n-1)d/2 = n[a + (n-1)d/2].

-a(m) = -a - (m-1)d < a(1) = a,

0 < 2a + (m-1)d = 2[a + (m-1)d/2].

s(m) = m[a + (m-1)d/2] >0.

a(1) = a < -a(m+1) = -a - md,

0 > 2a + md = 2[a + md/2]

s(m+1) = (m+1)[a + md/2] < 0.

答案：A.s(m)>0且s(m+1)