1.
an=a+(n-1)d
sn=na+n(n-1)d/2
s9=9a+36d=18,a+4d=2,
s(n-4)=(n-4)a+(n-4)(n-5)d/2=30
sn=na+n(n-1)d/2=240,
sn-s(n-4)=4a+(4n-10)d=240-30=210
2a+(2n-5)d=105
2(2-4d)+(2n-5)d=105
(2n-13)d=101
d=101/(2n-13)
a=2-4d=2-404/(2n-13)=(4n-430)/(2n-13)
s(n-4)=(n-4)a+(n-4)(n-5)d/2
=(n-4)(4n-430)/(2n-13)+202(n-4)(n-5)/(2n-13)
=30
sn=na+n(n-1)d/2
=n(4n-430)/(2n-13)+202n(n-1)/(2n-13)
=240
(n-4)(4n-430)+202(n-4)(n-5)=30(2n-13)
n(4n-430)+202n(n-1)=240(2n-13)
206n^2-2770n+5999=0
206n^2-1112n+3120=0
无解.
2.
设an=a+(n-1)d
sn=na+n(n-1)d/2
s10=10a+45d=140
a+4.5d=14
s奇=a1+a3+a5+a7+a9=5a+20d=125
a+4d=25
d=-22,a=113
an=113-22(n-1)=-22n+135
a6=-132+135=3;
3.若等差数列｛an｝的前4项和为25,后4项和为63,前n项和为286,求n
设an=a+(n-1)d
sn=na+n(n-1)d/2
s4=4a+6d=25
sn-s(n-4)=[na+n(n-1)d/2]-[(n-4)a+(n-4)(n-5)d/2]
=4a+(4n-10)d
=63
sn=na+n(n-1)d/2=286
所以
4a+6d=25
4a+(4n-10)d=63
na+n(n-1)d/2=286
用a=(25-6d)/4代入：
(2n-8)d=19
25n+(2n-8)dn=572
25n+19n=572
44n=572
n=13.