设数列{an}和{bn}都是等差数列,其中a1=25,b1=75,且a100+b100=100,则数列{an+bn}的前100项之和是
A.1000
B.10000
C.1100
D.11000
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B解析分析:由等差数列的性质可得{an+bn}是等差数列,且(a1+b1)+(a100+b100)=200,代入等差数列的前n项和公式即可求解.解答:∵{an}、{bn}都是等差数列,∴{an+bn}是等差数列,∵a1=25,b1=75,a100+b100=100,∴a1+b1+a100+b100=200,∴S100==10000故选B点评:本题综合考查等差数列的性质和前n项和公式,整体代入是解决问题的关键.属基础题.