已知等差数列{an}的公差d>0,设{an}的前n项和为Sn,a1=1,S2•S3=36.
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已知等差数列{an}的公差d>0,设{an}的前n项和为Sn,a1=1,S2•S3=36.
(Ⅰ)求d及Sn;
(Ⅱ)求m,k(m,k∈N*)的值,使得am+am+1+am+2+…+am+k=65.
(Ⅰ)求d及Sn;
(Ⅱ)求m,k(m,k∈N*)的值,使得am+am+1+am+2+…+am+k=65.
(Ⅰ)由a1=1,S2•S3=36得,
(a1+a2)(a1+a2+a3)=36,
即(2+d)(3+3d)=36,化为d2+3d-10=0,
解得d=2或-5,
又公差d>0,则d=2,
所以Sn=na1+
n(n−1)
2•d=n2(n∈N*).
(Ⅱ)由(Ⅰ)得,an=1+2(n-1)=2n-1,
由am+am+1+am+2+…+am+k=65得,
(k+1)(am+am+k)
2=65,
即(k+1)(2m+k-1)=65,
又m,k∈N*,则(k+1)(2m+k-1)=5×13,或(k+1)(2m+k-1)=1×65,
下面分类求
当k+1=5时,2m+k-1=13,解得k=4,m=5;
当k+1=13时,2m+k-1=5,解得k=12,m=-3,故舍去;
当k+1=1时,2m+k-1=65,解得k=0,故舍去;
当k+1=65时,2m+k-1=1,解得k=64,m=-31,故舍去;
综上得,k=4,m=5.
(a1+a2)(a1+a2+a3)=36,
即(2+d)(3+3d)=36,化为d2+3d-10=0,
解得d=2或-5,
又公差d>0,则d=2,
所以Sn=na1+
n(n−1)
2•d=n2(n∈N*).
(Ⅱ)由(Ⅰ)得,an=1+2(n-1)=2n-1,
由am+am+1+am+2+…+am+k=65得,
(k+1)(am+am+k)
2=65,
即(k+1)(2m+k-1)=65,
又m,k∈N*,则(k+1)(2m+k-1)=5×13,或(k+1)(2m+k-1)=1×65,
下面分类求
当k+1=5时,2m+k-1=13,解得k=4,m=5;
当k+1=13时,2m+k-1=5,解得k=12,m=-3,故舍去;
当k+1=1时,2m+k-1=65,解得k=0,故舍去;
当k+1=65时,2m+k-1=1,解得k=64,m=-31,故舍去;
综上得,k=4,m=5.
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