Question

设等差数列{an}的首项a1及公差d都为整数,前n项和为Sn,若a11=0,S14=98,求数列{an}的通项公式,并求

Answer 1

a11 = a1 + 10d = 0

S14 = 7(2a1 + 13d) = 98

解得：

a1 = 20

d = -2

所以 an = a1 + (n - 1)d = 20 - 2(n - 1) = -2n + 22

因为 a11 = 0 ,d < 0

当 n = 10 或 n = 11 时,Sn 最大 = 110