问题
填空题
已知等比数列的前n项和Sn=4n+a，则实数a=______．
答案
由于等比数列的前n项和Sn=4n+a，可得首项a1=S1=4+a，
a2=S2-S1=16+a-（4+a）=12，a3=S3-S2=64+a-（16+a）=48，
再由等比数列的定义可得
=12 4+a
=4，解得 a=-1，48 12
故答案为-1．
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已知等比数列的前n项和Sn=4n+a，则实数a=______．
由于等比数列的前n项和Sn=4n+a，可得首项a1=S1=4+a，
a2=S2-S1=16+a-（4+a）=12，a3=S3-S2=64+a-（16+a）=48，
再由等比数列的定义可得
=12 4+a
=4，解得 a=-1，48 12
故答案为-1．
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