等比数列ans3=6s6=9求s12等差数列an前n项和为sn等比数列bn前n项和为tnsn/tn=n/2n+1求a7/b7

Question

等比数列an s3=6 s6=9 求s12 等差数列an前n项和为sn 等比数列bn前n项和为tn sn/tn=n/2n+1 求a7/b7

Answer 1

a6=a1+a2+a3+a4+a5+a6
=(a1+a2+a3)+q³(a1+a2+a3)
=(1+q³)(a1+a2+a3)
=(1+q³)S3=6(1+q³)=9
1+q³=3/2
q³=1/2
S12=a1+a2+...+a6+a7+...+a12
=(a1+a2+...+a6)+q^6(a1+a2+...+a6)
=(1+q^6)(a1+a2+...+a6)
=[1+(q³)²]S6
=[1+(1/2)²]×9
=81/8
第二题估计是抄错了,{bn}也应该是等差数列.
a7/b7=S(2×7-1)/T(2×7-1)
=S13/T13
=13/(2×13+1)
=13/27