已知等差数列an的前n项和为Sn,a2=4,S10=110,则当Sn-an取最小值时,an=A.0 B.1 C.2 D.-1/4设a1=a,an=a+(n-1)da2=a+d=4S10=10a+10(10-1)d/2=110由以上可得a=d=2所以an=2+(n-1)×2=2nSn=na+n(n-1)d/2=n²+n则Sn-an=n²+n-2n=n²-n
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![已知等差数列an的前n项和为Sn,a2=4,S10=110,则当Sn-an取最小值时,an=A.0 B.1 C.2 D.-1/4设a1=a,an=a+(n-1)da2=a+d=4S10=10a+10(10-1)d/2=110由以上可得a=d=2所以an=2+(n-1)×2=2nSn=na+n(n-1)d/2=n²+n则Sn-an=n²+n-2n=n²-n](/uploads/image/z/4049798-14-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca2%3D4%2CS10%3D110%2C%E5%88%99%E5%BD%93Sn-an%E5%8F%96%E6%9C%80%E5%B0%8F%E5%80%BC%E6%97%B6%2Can%3DA.0+B.1+C.2+D.-1%2F4%E8%AE%BEa1%3Da%2Can%3Da%2B%28n-1%29da2%3Da%2Bd%3D4S10%3D10a%2B10%2810-1%29d%2F2%3D110%E7%94%B1%E4%BB%A5%E4%B8%8A%E5%8F%AF%E5%BE%97a%3Dd%3D2%E6%89%80%E4%BB%A5an%3D2%2B%28n-1%29%C3%972%3D2nSn%3Dna%2Bn%28n-1%29d%2F2%3Dn%26%23178%3B%2Bn%E5%88%99Sn-an%3Dn%26%23178%3B%2Bn-2n%3Dn%26%23178%3B-n)
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已知等差数列an的前n项和为Sn,a2=4,S10=110,则当Sn-an取最小值时,an=A.0 B.1 C.2 D.-1/4设a1=a,an=a+(n-1)da2=a+d=4S10=10a+10(10-1)d/2=110由以上可得a=d=2所以an=2+(n-1)×2=2nSn=na+n(n-1)d/2=n²+n则Sn-an=n²+n-2n=n²-n
已知等差数列an的前n项和为Sn,a2=4,S10=110,则当Sn-an取最小值时,an=
A.0 B.1 C.2 D.-1/4
设a1=a,an=a+(n-1)d
a2=a+d=4
S10=10a+10(10-1)d/2=110
由以上可得a=d=2
所以an=2+(n-1)×2=2n
Sn=na+n(n-1)d/2=n²+n
则Sn-an=n²+n-2n=n²-n
所以当x为1/2时,n²-n最小为-1/4
此时an=2n等于2×1/2等于1.
可是答案是2.
已知等差数列an的前n项和为Sn,a2=4,S10=110,则当Sn-an取最小值时,an=A.0 B.1 C.2 D.-1/4设a1=a,an=a+(n-1)da2=a+d=4S10=10a+10(10-1)d/2=110由以上可得a=d=2所以an=2+(n-1)×2=2nSn=na+n(n-1)d/2=n²+n则Sn-an=n²+n-2n=n²-n
楼主前面的步骤都没有算错,
(前面步骤和你算的一样我就不写了)
但是在则Sn-an=n²+n-2n=n²-n中
∵n≥1
∴N不能取1/2
∴满足条件的n的最小值是 n=1
将n=1带回原式
an=2
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