设等差数列{an}的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1
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设等差数列{an}的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1
设等差数列{an}的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1
设等差数列{an}的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1
S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=3,a1=-20
∴an=a1+(n-1)d=3n-23
由an>0即3n-23>0得 n>23/3
所以当n0;
设Tn为{|an|}的前n项和
当n=8时,Tn=|a1|+|a2|+|a3|+…+|a7|+|a8|+…+|an|
=-a1-a2-a3-…-a7+a8+…+an
=Sn-2S7
再代入计算就行了
S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=-28,a1=57.5
∴an=a1+(n-1)d=57.5+(n-1)x(-28)=-28n+85.5
由-28n+85.5>0且n为正整数得n=1,2,3
所以,当n小于等于3时,︳a1︳+︳a2︳+︳a3︳+…+︳an|=na1+n(n-1)d=57.5n+n(n-1)...
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S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=-28,a1=57.5
∴an=a1+(n-1)d=57.5+(n-1)x(-28)=-28n+85.5
由-28n+85.5>0且n为正整数得n=1,2,3
所以,当n小于等于3时,︳a1︳+︳a2︳+︳a3︳+…+︳an|=na1+n(n-1)d=57.5n+n(n-1)x(-28)
当n大于3时,︳a1︳+︳a2︳+︳a3︳+…+︳an|=-sn+2s3=....
(后面不写了,应该能看懂,能知道后面的应该如何做了)
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S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=3,a1=-20
∴an=a1+(n-1)d=3n-23