数列{an}对任意n∈N*,满足an+1=an+3,且a3=8,则S10等于
A.155
B.160
C.172
D.240
网友回答
A解析分析:由an+1=an+3,知{an}是公差d=3的等差数列,由a3=a1+2d=8,知a1=8-2d=8-6=2,由此能求出S10的值.解答:∵an+1=an+3,∴an+1-an=3,∴{an}是公差d=3的等差数列,∵a3=a1+2d=8,∴a1=8-2d=8-6=2,∴=10×2+5×9×3=155.故选A.点评:本题考查等差数列的前n项和的计算,是基础题.解题时要认真审题,仔细解答.