> 数学 >
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Sn的最大
人气:104 ℃ 时间:2019-10-23 07:44:15
解答
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
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版