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等差数列{an}满足a=5,a4+a10=30,{an}的前n项和为Sn求an及Sn
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等差数列{an}满足a=5,a4+a10=30,{an}的前n项和为Sn 求an及Sn
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答案和解析
an=a1+(n-1)d
a5=5
a1+4d=5 (1)
a4+a10=30
a1+6d=15 (2)
(2)-(1)
d=5
from (1) => a1=-15
an =-15+5(n-1) =5n-20
Sn = a1+a2+...+an
= 5(n -7)n/2
a5=5
a1+4d=5 (1)
a4+a10=30
a1+6d=15 (2)
(2)-(1)
d=5
from (1) => a1=-15
an =-15+5(n-1) =5n-20
Sn = a1+a2+...+an
= 5(n -7)n/2
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