(2014. The three module in Hexi District) set Sn as the first n term and the 8a2+a5=0 of the geometric sequence {an}, then S5S2 equals (). A. 11B. 5C. -8D. -11

(2014. The three module in Hexi District) set Sn as the first n term and the 8a2+a5=0 of the geometric sequence {an}, then S5S2 equals (). A. 11B. 5C. -8D. -11

Let the common ratio of the equal ratio sequence {an} be q, (Q ≠ 0) from the meaning of the title, 8a2 + A5 = 8a1q + a1q4 = 0, the solution is q = - 2, so s5s2 = A1 (1 − Q5) 1 − QA1 (1 − Q2) 1 − q = 1 − q51 − Q2 = 1 − (− 2) 51 − (− 2) 2 = - 11, so D is selected

(2014. The three module in Hexi District) set Sn as the first n term and the 8a2+a5=0 of the geometric sequence {an}, then S5S2 equals ().
A. 11B. 5C. -8D. -11

Let the common ratio of the equal ratio sequence {an} be q, (Q ≠ 0) from the meaning of the title, 8a2 + A5 = 8a1q + a1q4 = 0, the solution is q = - 2, so s5s2 = A1 (1 − Q5) 1 − QA1 (1 − Q2) 1 − q = 1 − q51 − Q2 = 1 − (− 2) 51 − (− 2) 2 = - 11, so D is selected

It is known that SN is the sum of the first n terms of the proportional sequence {an}, 8a2 + A5 = 0, then S5 / S2 =?
Please explain the detailed solution of each step,

Let the common ratio be q, the first item A1 and the common ratio Q are not equal to 0
8a2+a5=0
8a1q+a1q^4=0
A1 is not equal to 0, both sides of the equation divide by A1
q^4+8q=0
q(q^3+8)=0
q(q+2)(q^2-2q+4)=0
Q is not equal to 0, Q ^ 2-2q + 4 = (Q-1) ^ 2 + 3 constant > 0, if the equation holds, only Q + 2 = 0
q=-2
S5/S2=[a1(q^5 -1)/(q-1)]/[a1(q^2 -1)/(q-1)]
=(q^5 -1)/(q^2 -1)
=(-32-1)/(4-1)
=-33/3
=-11

If SN is the sum of the first n terms of the equal ratio sequence {an}, 8a2 + A5 = 0, then s6s3=______ ．

From 8a2 + A5 = 0, we get that a5a2 = Q3 = - 8s6s3 = A1 (1 − Q6) 1 − QA1 (1 − Q3) 1 − q = 1 − q61 − Q3 = - 7, so the answer is: - 7