差数列,故S9=9a1+×=9+18=27.]
8.已知等差数列{an}的前n项和为Sn,且6S5-5S3=5,则a4=________.
【导学号:91432172】
[设等差数列{an}的首项为a1,公差为d,由6S5-5S3=5,得3(a1+3d)=1,所以a4=.]
三、解答题
9.等差数列{an}中,a10=30,a20=50.
(1)求数列的通项公式;
(2)若Sn=242,求n.
[解] (1)设数列{an}的首项为a1,公差为d.
则解得
∴an=a1+(n-1)d=12+(n-1)×2=10+2n.
(2)由Sn=na1+d以及a1=12,d=2,Sn=242,
得方程242=12n+×2,即n2+11n-242=0,解得n=11或n=-22(舍去).故n=11.
10.已知等差数列{an}的前n项和Sn=n2-2n,求a2+a3-a4+a5+a6.
【导学号:91432173】
[解] ∵Sn=n2-2n,
∴当n≥2时,an=Sn-Sn-1
=n2-2n-[(n-1)2-2(n-1)]
=n2-2n-(n-1)2+2(n-1)
=2n-3,