(2014•淮安模拟)设数列{an}为等差数列,数列{bn}为等比数列.若a1<a2,b1<b2,且bi=ai2(i=1
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(2014•淮安模拟)设数列{an}为等差数列,数列{bn}为等比数列.若a1<a2,b1<b2,且bi=ai2(i=1,2,3),则数列{bn}的公比为
3+2
2 |
设等差数列{an}的公差为d,
由a1<a2可得d>0,
∴b1=a12,b2=a22=(a1+d)2,
b3=a32=(a1+2d)2,
∵数列{bn}为等比数列,∴b22=b1•b3,
即(a1+d)4=a12•(a1+2d)2,
∴(a1+d)2=a1•(a1+2d) ①
或(a1+d)2=-a1•(a1+2d),②
由①可得d=0与d>0矛盾,应舍去;
由②可得a1=
−2−
2
2d,或a1=
−2+
2
2d,
当a1=
−2−
2
2d时,可得b1=a12=
3+2
2
2d2
b2=a22=(a1+d)2=
1
2d2,此时显然与b1<b2矛盾,舍去;
当a1=
−2+
2
2d时,可得b1=a12=
3−2
2
由a1<a2可得d>0,
∴b1=a12,b2=a22=(a1+d)2,
b3=a32=(a1+2d)2,
∵数列{bn}为等比数列,∴b22=b1•b3,
即(a1+d)4=a12•(a1+2d)2,
∴(a1+d)2=a1•(a1+2d) ①
或(a1+d)2=-a1•(a1+2d),②
由①可得d=0与d>0矛盾,应舍去;
由②可得a1=
−2−
2
2d,或a1=
−2+
2
2d,
当a1=
−2−
2
2d时,可得b1=a12=
3+2
2
2d2
b2=a22=(a1+d)2=
1
2d2,此时显然与b1<b2矛盾,舍去;
当a1=
−2+
2
2d时,可得b1=a12=
3−2
2
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