文档介绍：等差数列?知识梳理?课前热身??课时小结数列中从第二项起,每一项与它前一项的差等于一个常数,这个数列叫做等差数列。:为常数)ddaann(1???:dnaan)1(1???:2)(1nnaanS??:等差数列,若m+n=p+q,则qpnmaaaa??+??nadnnnaSn2)1(1??? 等差数列?课前热身1 . 已知等差数列{an},a1=1 ,d=2, 求a201 . 2 . 已知等差数列{an},a5=-1 ,a8=2,求a1与d. 解:a201=a1+(n-1)d =1+(201-1)×2 =401. 解:a5 = a1+(5-1)d = -1a8 = a1+(8-1)d = 2a1 = -5, d= 等差数列?课前热身dnnnaSn2)1(1???3 . 已知等差数列{an},a1=1 ,d =-2,求S20 . 解:36022)120(2012020???????)(-S4()a1=-7,且满足an+1=an+2(n∈N),则a1+a2+a3+…+a17=_____ . 等差数列解:由an+1=an+2可知{an})117(17)7(172)1(117=?????????dnnnaS4()a1=-7,且满足an+1=an+2(n∈N),则a1+a2+a3+…+a17=_____ . ? 等差数列例1. 已知等差数列{ an },a4=9 ,a9=-6 , Sn=63 . 求n . ?:a4 = a1+(4-1)d = 9a9 = a1+(9-1)d = -6得a1 = 18, d=-)1(18??????)(- 等差数列可求得n1 = 6 或n2= :1.()已知等差数列{an},a1= a2+a5=4,an=33,则n为().31(A) 48 (B) 49 (C) 50 (D) 等差数列解:a2 + a5 =(a1+d)+(a1+4d)=2a1+5d=4an = a1+(n-1)d = 33可求得n= ??d32?d3332)1(31????nC? . 已知等差数列{ a n },a3+a15= + a11 ; :等差数列{ a n }中, a7 + a11 =a3+a15 =30; 同理a9 + a 9 = a 3 + a15 =30 所以a9 = 等差数列练****1.()已知等差数列{ a n },a7+a9=16. a4 =1,求a12 ; { a n },a3+a4+a5+a6+a7=150. 求a2 + a8 .{ a n },a1+a2+a3+a4+a5=20. 求a3 ; 等差数列例3.()等差数列{an}中,a1+a2+a3=-24,a18+a19+a20=78,则此数列前20项和等于()(A) 160 (B) 180 (C) 200 (D) 220 ?)(202)(201120aaaanSn????解:等差数列{a n}, a1+a2+a3+a18+a19+a20=54 3(a1+a20)=54 a1+a20=18 18021820??? 等差数列