3．选B．?a1?a3?2a2?a2?5,a1a3?16，?d?3,a1?2，?a11+a12+a13＝３a12＝105
4．填３0.∵S偶-S奇=5d，∴d?5,a1?5∴a6?30. 或S偶?a2?a4?a6?a8?a10?5a6?150 ∴a6?30．
?a3?a1?2d?165．填110.??，?a1?20,d??2，?S10?110．
S?20a?190d?201?206．填50．?a1?9,d??2,令an?11?2n?0?n?5.5，a1?a2?a3???a10
?a1???a5?a6???a10?S5?S10?S5?25?0?25?50
课时训练
1．选D.∵an+1 －an = 2∴数列{an}是公差为2的等差数列,所以a51 =101. 2．选D.∵an＝－2n＋1,∴a1＝－1,d??2,∴Sn?(a1?an)?nS??n2,?n??n
2nSn数列{}是首项为?1，公差为?1的等差数列，前11项和为－66．
n
9(a1＋a9)
2S99a595
3．选A．由已知得：＝＝＝×＝1.
S55(a1＋a5)5a359
2
4．选A．∵数列{an}，{bn}是等差数列，数列?an?bn?是等差数列，首项为a1＋b1＝4，公差为１，?a37?b37?4?(37?1)?1?40
?a1?1,d?2，5．选D．∴Sk?2?Sk?ak?2?ak?1?a2??2k4??4241?k4，k?5．
111111
6．选B．数列{}是等差数列，令bn?，则b3?? ?，b7?an＋1a7?12an?1a3?13b7?b311221，?b11?b3?(11?3)d?，?b11??，?a11? ?a11?137?324327．解：?S3,S6?S3,S9?S6成等差数列，?S9?63． ?d?8．解：∵a2是－8和10的等差中项，∴a2＝1. 9．解:∵an= a5+(n－5)d，∴a12＝a5+7d >31∴d >3.
11－319S11a1＋a112a62×a619
∵＝＝＝＝，∴＝. T11b1＋b112b64×b64111－34110．解：?12a1?66d?8(4a1?6d)，?a19? d1011．(1)设等差数列的首项为a1，公差为d，∵a5＝15，∴a1＋4d＝15，① ∵a10＝25，∴a1＋9d＝25，② 解①②组成的方程组得：a1＝7，d＝2. ∴an＝7＋(n－1)×2＝2n＋5.
11
1
(2)∵Sn＝112，∴7n＋n(n－1)×2＝112.即：n2＋6n－112＝0，解之得n＝－14(舍去)
2或n＝8，故n＝8.
12．解:?a1?a2?a3?a4?40，an?an?1?an?2?an?3?80， 又?a1?an?a2?an?1?a3?an?2?a4?an?3，?a1?an?30?Sn?(a1?an)n?210 2?n?14
13．解：（１）∵bn?22，∴bn?1?， an?1an?1?12222∴bn?bn?1?????2∴数列{bn}是公差为２的等差
1an?1an?1?12??1an?1?1an?1222n?5数列，b1? ??5，?bn?2n?7??an?a1?1an?12n?7n(n?1)（２）?Tn??5n??2?n2?6n?(n?3)2?9，?n?3时Tn取最小值
212