2022英语周报七年级新目标的答案

17.解:本题考查数列递推关系与通项公式因为a2 3an=6Sn b,所以a6S 1 b,两式相减得3(即(3)(>0,所以a4分若选择①,因为a3=10,且a3-a1=2×3,所以a1=46Sn b,得6a1 b,即b把a1=4代入上式,得b当b=4时,由a-3a1-4=0及a1>0,得a1=4.所以存在实数b=4,使得数列{an}是以4为首项,以3为公差的等差数列,数列{an}的通项公式为an=4 3(n若选择②,由an 1-an=3,得a2-a1=3,又已知S2=a1 a2=9,所以a1=3,a由a2 3an=6Sn b,得3a1,所以b当b=0时,由ai-3a1=0及a1>0,得a1=3由a23及所以a1=3和a2=6满足an=3.所以存在实数b=0,使得数列{an}是以3为首项,以3为公差的等差数列,数列{an}的通项公式为an=3 3(n-1)=310分若选择③由a2 3an=6Sn b,得ai 3a1=6a1 b,即因为已知数列{an}的各项均为正数,所以a1>0因为关于a1的一元二次方程ai-3a1-b=0至少存在一个正实数解的充要条件是△=9 4b≥0,解得b≥这与已知条件b<-3矛盾,所以不存在实数b使得数列{an}是等差数列分注:若x2-3x-b=0存在两个实数解分别为x1,x2,则x1 x(注:若x2-3x-b=0存在两个实数解分别为x1,x2,则x1 x2=3,x1x当b>0时,x2-3x-b=0的解一正一负;当b=0时,x2-3x-b=0的解一正一零;当一9≤6<0时,x3x-b=0的解均为正.所以当且仅当△=9 46≥0时,方程x2-3x-b=0至少存在一个正实数解.)

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