设等差数列{an}的前n项的和为Sn,且S4=-62,S6=-75求(1){an}的通项公式an及前n项的和Sn(2)|a1|+|a2|+|a3|+……+|a14|=
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![设等差数列{an}的前n项的和为Sn,且S4=-62,S6=-75求(1){an}的通项公式an及前n项的和Sn(2)|a1|+|a2|+|a3|+……+|a14|=](/uploads/image/z/677027-11-7.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8C%E4%B8%BASn%2C%E4%B8%94S4%3D-62%2CS6%3D-75%E6%B1%82%EF%BC%881%EF%BC%89%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%E5%8F%8A%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8CSn%EF%BC%882%EF%BC%89%7Ca1%7C%2B%7Ca2%7C%2B%7Ca3%7C%2B%E2%80%A6%E2%80%A6%2B%7Ca14%7C%3D)
设等差数列{an}的前n项的和为Sn,且S4=-62,S6=-75求(1){an}的通项公式an及前n项的和Sn(2)|a1|+|a2|+|a3|+……+|a14|=
设等差数列{an}的前n项的和为Sn,且S4=-62,S6=-75求(1){an}的通项公式an及前n项的和Sn
(2)|a1|+|a2|+|a3|+……+|a14|=
设等差数列{an}的前n项的和为Sn,且S4=-62,S6=-75求(1){an}的通项公式an及前n项的和Sn(2)|a1|+|a2|+|a3|+……+|a14|=
4a1+d+2d+3d=-62;4a1+6d=-62;
6a1+d+2d+3d+4d+5d=-75,6a1+15d=-75;
a1=-20;d=3;
(1)an=a1+3(n-1)=3n-3-20=3n-23;
Sn=3*1-23+3*2-23+...+3*n-23=3(1+2+...+n)-23n=3n(n+1)/2-23n;
(2)|a1|+|a2|+|a3|+……+|a14|=20+17+14+11+8+5+2+1+4+7+10+13+16+19=25*3+20*3+10=75+70=145;
(1)设等差数列an=a1+(n-1)d,Sn=na1+n(n-1)d/2,则S4=2(a1+a4)=4a1+6d,S6=3(a1+a6)=6a1+15d,那么,S6+S4=10a1+21d=‐137,S6‐S4=2a1+9d=10a1+45d=‐13*5=‐65
解得d=3,则a1=‐20,所以an=‐20+3(n-1)
(2)由通项公式得a7=‐2,a8=1,则有|...
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(1)设等差数列an=a1+(n-1)d,Sn=na1+n(n-1)d/2,则S4=2(a1+a4)=4a1+6d,S6=3(a1+a6)=6a1+15d,那么,S6+S4=10a1+21d=‐137,S6‐S4=2a1+9d=10a1+45d=‐13*5=‐65
解得d=3,则a1=‐20,所以an=‐20+3(n-1)
(2)由通项公式得a7=‐2,a8=1,则有|a1|+a8=|a2|+a9=······=|a7|+a14=21,所以,
|a1|+|a2|+|a3|+······+|a14|=21*7=147
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