已知等差数列{a n }的前n项和为S n ,a 4 +a 7 +a 10 =9,S 14 -S 3 =77,则使S n 取得最小值时n的值为(

2024-11-14 13:51:37
推荐回答(1个)
回答1:

等差数列{a n }中,
∵a 4 +a 7 +a 10 =9,S 14 -S 3 =77,
a 7 = a 1 +6d=3
S 14 - S 3 =(14 a 1 +
14×13
2
d)-(3 a 1 +
3×2
2
d)=77

解得a 1 =-9,d=2.
S n =-9n+
n(n-1)
2
×2

=n 2 -10n
=(n-5) 2 -25,
∴当n=5时,S n 取得最小值.
故选B.