Note that the sum of the first n terms of the equal ratio sequence {an} is Sn, if S3 = 2, S6 = 18, then s10s5 equals () A. -3B. 5C. -31D. 33

Note that the sum of the first n terms of the equal ratio sequence {an} is Sn, if S3 = 2, S6 = 18, then s10s5 equals () A. -3B. 5C. -31D. 33

According to the title, S3 = 2, S6 = 18, easy to get Q ≠ 1; ∵ S3 = 2, S6 = 18, ∵ s3s6 = A1 (1 − Q3) 1 − QA1 (1 − Q6) & nbsp; 1 − q = 1 − Q31 − Q6 = 11 + Q3 = 218, ∵ q = 2. ∵ s10s5 = A1 (1 − Q10) & nbsp; 1 − QA1 (1 − Q5) 1 − q = 1 − Q101 − Q5 = 1 + Q5 = 33

Let the common ratio Q of the equal ratio sequence an = 2 and the sum of the first n terms be Sn, then what is the value of S3 of A3

Formula:
an=a1*q^(n-1)
sn=a1*(1-q^n)/(1-q)
a3/s3=a1*2^(3-1) *(1-2)/[a1*(1-2^3)]=4/7
I was wrong just now. It should be
s3/a3=7/4