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由a2+a3+a4=b4可得到3*a3=b4;
a3=9
b4=27
a3=9
b4=27
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an = a1+(n-1)d
bn=b1.q^(n-1)
a1=b1=1
a2+a3+a4 =b4
3a1+6d = b1.q^3
3+6d =q^3 (1)
(b4)^2 = 81a3
(b1.q^3)^2 = 81(a1+2d)
q^6 = 81(1+2d)
(3+6d)^2 =81(1+2d)
(1+2d)^2 =9(1+2d)
4d^2 -14d - 8 =0
2d^2-7d-4=0
(d-4)(2d+1)=0
d=4 or -1/2
case 1: d=4
a3 = a1+2d = 1+ 8 = 9
b4 = q^3 =3+6d = 3+24 =27
case 2: d=-1/2
a3 = a1+2d = 1-1 =0
b4 = q^3 =3+6d = 3-3 =0
bn=b1.q^(n-1)
a1=b1=1
a2+a3+a4 =b4
3a1+6d = b1.q^3
3+6d =q^3 (1)
(b4)^2 = 81a3
(b1.q^3)^2 = 81(a1+2d)
q^6 = 81(1+2d)
(3+6d)^2 =81(1+2d)
(1+2d)^2 =9(1+2d)
4d^2 -14d - 8 =0
2d^2-7d-4=0
(d-4)(2d+1)=0
d=4 or -1/2
case 1: d=4
a3 = a1+2d = 1+ 8 = 9
b4 = q^3 =3+6d = 3+24 =27
case 2: d=-1/2
a3 = a1+2d = 1-1 =0
b4 = q^3 =3+6d = 3-3 =0
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