等差数列{an}3a4=7a7,且a1>0sn是等差数列{an}的前n项和,若sn取最大值,

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等差数列{an}3a4=7a7,且a1>0sn是等差数列{an}的前n项和,若sn取最大值,
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等差数列{an}3a4=7a7,且a1>0sn是等差数列{an}的前n项和,若sn取最大值,
等差数列{an}3a4=7a7,且a1>0sn是等差数列{an}的前n项和,若sn取最大值,

等差数列{an}3a4=7a7,且a1>0sn是等差数列{an}的前n项和,若sn取最大值,
设公差为d
所以
a7=a4+3d

3a4=7a7
7(a1+6d)=3(a1+3d)
所以 a1=-33/4d
因为sn取最大
所以a1>0
所以啊an=-33/4d+(n-1)d
an=nd-37/4d
数列中最小的正数即为解
所以
n=9

由题意可知:3(a1+3d)=7(a1+6d)
则有: d=-4/33 a1
那么有:an =-4/33 a1(n-1)+a1
令an>o,即可解得: n<37/4
取整,则n=8