It is known that SN is the sum of the first n terms of the equal ratio sequence {an}, S2, S6 and S4 are equal difference sequence, and the common ratio Q of the sequence {an} is obtained?

Since the sequence {an} is an equal ratio sequence, then an = A1 * q ^ (n-1),
1) If q = 1, S2 = 2A1, S6 = 6A1, S4 = 4A1
S2, S6 and S4 are equal difference sequence, then S2 + S4 = 2s6, thus A1 = 0, which does not meet the condition of equal ratio sequence, so it is discarded;
2) If Q ≠ 1, Sn = A1 * (1-Q ^ n) / (1-Q),
So S2 = A1 * (1-Q ^ 2) / (1-Q), S6 = A1 * (1-Q ^ 6) / (1-Q), S4 = A1 * (1-Q ^ 4) / (1-Q),
Because S2, S6 and S4 are arithmetic sequences,
So (S2 + S4) = 2s6, that is, A1 * (1-Q ^ 2) / (1-Q) + A1 * (1-Q ^ 4) / (1-Q) = A1 * (1-Q ^ 6) / (1-Q),
It is concluded that Q ^ 2 * (Q ^ 2-1) (2q ^ 2 + 1) = 0
The solution is: q = 0 or q = ± 1
But when q = 0 or 1, they are not consistent, so they are discarded
So q = - 1

If S2 = 6, S4 = 30, then S6=______ ．

According to the properties of the equal ratio sequence, it is obtained that S2, s4-s2, s6-s4 are equal ratio sequence, that is, (s4-s2) 2 = S2 (s6-s4), S2 = 6, S4 = 30, substituting into: (30-6) 2 = 6 (s6-30), the solution is S6 = 126

If S2 = 2, S4 = 10, then what is S6 equal to

Sn, s2n Sn, s3n-s2n in the equal ratio sequence are also equal ratio (see the proof below)
S4-S2=10-2=8
S2×(S6-S4)=(S4-S2)^2
2×(S6-10)=8^2
S6-10=32
S6=42
Sn=A1+A2+…… +An
S2n-Sn=A(n+1)+A(n+2)+…… +A2n=(A1+A2+…… +An)×q^n=Sn×q^n
S3n-S2n=A(2n+1)+A(2n+2)+…… +A3n=(A1+A2+…… +An)×q^(2n)=Sn×(q^n)^2
Sn, s2n Sn, s3n-s3n are also equal

It is known that SN is the sum of the first n terms of the equal ratio sequence {an}, S2, S6 and S4 are equal difference sequence, and the common ratio Q of the sequence {an} is obtained?