设等比数列{a n}的公比为q,(q>0)
由题意可得2×
1
2 a 3 =3a 1+2a 2,
即 a 1 q 2 =3 a 1 +2 a 1 q ,即q 2-2q-3=0
解之可得q=3,或q=-1(舍去)
故
S 11 - S 9
S 7 - S 5 =
a 10 + a 11
a 6 + a 7 =
a 6 q 4 + a 7 q 4
a 6 + a 7 =q 4=81
故答案为:81
设等比数列{a n}的公比为q,(q>0)
由题意可得2×
1
2 a 3 =3a 1+2a 2,
即 a 1 q 2 =3 a 1 +2 a 1 q ,即q 2-2q-3=0
解之可得q=3,或q=-1(舍去)
故
S 11 - S 9
S 7 - S 5 =
a 10 + a 11
a 6 + a 7 =
a 6 q 4 + a 7 q 4
a 6 + a 7 =q 4=81
故答案为:81