问题
选择题
已知等差数列{an}的前n项和为Sn,且S2=10,S5=55,则过点P(n,
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答案
S2=2a1+d=10
S5=5a1+10d=55,
解得d=4,a1=3,
Sn=3n+
×4=2n2+n,n(n-1) 2
kPQ=
=2,
-Sn+2 n+2 Sn n n+2-n
∴直线PQ的方程为:y-
=2(x-n),Sn n
解得y=2x+1.
故选A.
已知等差数列{an}的前n项和为Sn,且S2=10,S5=55,则过点P(n,
|
S2=2a1+d=10
S5=5a1+10d=55,
解得d=4,a1=3,
Sn=3n+
×4=2n2+n,n(n-1) 2
kPQ=
=2,
-Sn+2 n+2 Sn n n+2-n
∴直线PQ的方程为:y-
=2(x-n),Sn n
解得y=2x+1.
故选A.