题目内容
已知{an}为等差数列,a15=8,a60=20,则a75=______________.
24
解法一:由a15=a1+14d=8,a60=a1+59d=20,
解得a1=
,d=
,∴a75=a1+74d=24.
解法二:a15,a30,a45,a60,a75成等差数列,设公差为d1,则d1=
=4,∴a75=a60+d1=24.
解法三:a60=a15+(60-15)d,∴d=
.∴a75=a60+(75-60)d=24.
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