已知等差数列{an}中:(1)a1=3⼀2,d=-1⼀2,Sn=-15,求n及a12 (2)S10

2025-06-22 18:11:06
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回答1:

(1).

-15=n(3/2)+n(n--1)(-1/2)/2
-60=6n-n^2+n
n^2-7n-60=0
(n-12)(n+5)=0
n=12
(2)
100=10(a1+a10)/2
a1+a10=20
a4+a7=a1+a10=20