Sn是等差数列{an}的前n项和,已知S9=18,Sn=240,an-4=30(n>9),求n
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Sn是等差数列{an}的前n项和,已知S9=18,Sn=240,an-4=30(n>9),求nSn是等差数列{an}的前n项和,已知S9=18,Sn=240,an-4=30(n>9),求nSn是等差数
Sn是等差数列{an}的前n项和,已知S9=18,Sn=240,an-4=30(n>9),求n
Sn是等差数列{an}的前n项和,已知S9=18,Sn=240,an-4=30(n>9),求n
Sn是等差数列{an}的前n项和,已知S9=18,Sn=240,an-4=30(n>9),求n
s9=(a1+a9)*9/2=(a1+a1+8d)*9/2=9a1+36d 即 9a1+36d=18 则 a1+4d=2 ①
a(n-4)=a1+(n-5)d 则 a1+(n-5)d=30 ②
由 ②-① 得 (n-9)d=28 ③
sn=(a1+an)*n/2=[a1+a1+(n-1)d]*n/2=[2a1+(n-9)d+8d]*n/2=[2a1+28+8d]*n/2=[a1+4d+14]*n=[2+14]*n=16n
则 16n=240 n=15
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