数学的等比数列问题求解答:两解:1.a1=1,q=2,an=2(n-1)次方 2.a1=4,q=1/2,an=4*(1/2)的(n-1)次方=8/2的n次方.过程如下:a1*a2*a3=8,即:a1q=2,代入a1+a2+a3=7并化简:2q方-5q+2=0,q=2或1/2,a1=1,或4,
设等比数列{an}公比为q,a1不等于0,前n项和为sn,若s3,s9,s6成等差数列...答:解:(1)若q=1,则S3 =3a ,S9 =9a ,S6 =6a;不成等差数列 故q≠1,此时由S3 , S9 S6 成等差数列得 2S9 = S3 + S6 ,2*a1(1-q^9)/(1-q)=a1(1-q)^3/(1-q)+a1(1-q^6)/(1-q)化简得:2q^6-q^3-1=0 所以q=- (1/2)^(1/3 ...