It is known that sn is the sum of the first n terms of the proportional series, and S3, S9, and S6 a

1、s3=a1*(1-q^3)/(1-q) s9=a1*(1-q^9)/(1-q) s6=a1*（1-q^6)/(1-q)

S3, S9, S6 into a series of equal differences, then 2S9 = S3 + S6, i.e.

2*a1*(1-q^9)/(1-q)=a1*(1-q^3)/(1-q)+a1*(1-q^6)/(1-q)

Simplified: q3(2q+1)(q-1)=0

0 and 1 are both root additions, and if they are rounded, then: q=(-1 2) (1 3).

2、ak=a1*[(1/2)^(1/3)]^k-1)=a1*(-1/2)^[k-1)/3]

a(k+6)=a1*[-1/2)^(1/3)]^k+5)=a1*(-1/2)^[k+5)/3]

a(k+3)=a1*[(1/2)^(1/3)]^k+2)=a1*(-1/2)^[k+2)/3]

ak+a(k+3)=a1*(-1/2)^[k-1)/3]+a1*(-1/2)^[k+2)/3]

a1*(-1/2)^[k-1)/3]*[1+(-1/2)^(3/3)]

a1*(-1/2)^[k-1)/3]/2

2*a(k+6)=2*a1*(-1/2)^[k+5)/3]

a1*(-1/2)^[k+5)/3]*(1/2)^(2)/2

a1*(-1/2)^[k+5)/3-2]/2

a1*(-1/2)^[k-1)/3]/2

That is, ak+a(k+3)=2*a(k+6).

Therefore, ak, ak+6, and ak+3 form a series of equal differences.

s3=a1(1-q) slam (1-q),s9=a1(1-q 9) (1-q),s6=a1(1-q 6) (1-q),2s9=s3+s6, 2a1(1-q 9) (1-q)=a1(1-q laughing) (1-q)+a1(1-q 6) (1-q),2q 9=q +q 6,2q 6=1+q , (q -1) (2q +1)=0 q =1 or q =-1 2, the common ratio of an is q=-(1 2) (1 3).

S2, S6, S4 are equal and the difference is a series.

2s6=s2+s4

2(1-q^6)=1-q^2+1-q^4

That is, the destruction of coarse 2q 6 = q 2 + q 4

q is 0 early, so 2q 2-q-1=0

q-1)(2q+1)=0

q = 1 or q 1 2

When Q=1, there are S3=3A1, S6=6A1, S9=9A1 meet the problem, so Q=1When Q≠1 is investigated, A1(1-Q 3) (1-Q)+A1(1-Q 6) (1-Q)=A1(1-Q 9) (1-Q) From Q≠1, remove the denominator to get (1-Q 3)+(1-Q 6)=(1-Q 9) so that Q 3=X, then (1-X)+(1-X 2)-(1-X 3)=.

s9-s6=(s6-s3)*q s6-s3=s3*q Use these two formulas to eliminate the noise and remove the square of s3*s6=s9

The first item is A1 per mu, and the ratio is Q

s6=s3+s3*q^3

s9=s3+s3*q^3+s3*q^6

S3, S9, S6 are equal to the series of vertical differences, then S3+S9=2S6, that is, S3+S9-2S6=0

s3+s3+s3*q^3+s3*q^6-2s3-2s3*q^3=s3*q^6-s3*q^3=0

s3*q^3(q^3-1)=0

q 3=1, so the common ratio of the proportional series is 1The tolerance of the equal difference series is respectful s3

When q=1, there are s3=3a1, s6=6a1, s9=9a1 that meet the problem, so q=1

When q≠1, a1(1-q 3) (1-q)+a1(1-q 6) (1-q)=a1(1-q 9) (1-q) is sorted by q≠1, denominator.

1-q 3) + (1-q 6) = (1-q 9) so that q 3 = x, then.

1-x）+（1-x^2）-（1-x^3)=01-x-x^2+x^3 =0

1-x)-x^2(1-x)=0

1-x)(1-x)(1+x)=0

1-x) 2(1+x)=0 is ≠1,x≠1, so x=-1, i.e. q 3=-1, q=-1

In summary, q = -1 or +1

Solve the ridge: the judgment base is infiltrated by the topic, s9-s3=s6-s9. s9-s3=a4+..a9

s6-s9=-(a7+a8+a9)

and (a4+a5+a6)+2(a7+a8+a9)=0a3(q+q+q)+2a6(q+q+q)=0a3=-2a6

So a6=-2a9

a3=4a9

Therefore, a3 + a6 = 2a9