设等比数列{an}的前n项和为Sn,若S4=3,S8=15,则a11+a12+a13+a14=

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2025-04-02 14:06:57
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回答1:

解由等比数列{an}的前n项和为Sn,
知sm,s2m-sm,s3m-s2m,s4m-s3m成等比数列
则由
S4=3
S8-S4=15-3=12
S12-S8=12×3=36
S16-S12=36×3=108
故a11+a12+a13+a14
=S16-S12
=108.