等差数列an的前n项和为sn已知s1,根号二s2,3s3成公比为q的等差数列
问题描述：
等差数列an的前n项和为sn已知s1,根号二s2,3s3成公比为q的等差数列
则q=
答
2S2=S1s3
2(2a1+d)^2=3a1(3a1+3d)
8a1^2+8a1d+2d^2=9a1^2+9a1d
a1^2+a1d-2d^2=0
(a1+2d)(a1-d)=0
a1=d或a1=-2d
当a1=d 时
s1=a1
√2S2=√2(2a1+d)=3√2a1
3s3=3(3a1+3d)=9a1
q=3√2
当a1=－2d 时.d=-a1/2
s1=a1
√2S2=√2(a1-a1/2=√2a1/2
3s3=3(3a1-3d/2)=9a1/2
不合题意
综上得：q=3√2倒数第四行 根号2s2应该=更好2（a1+a1-a1/2）我觉得a1=-2d是可取的