Question

3．(2011年高考四川卷)数列{an}的首项为3,{bn}为等差数列且bn＝an＋1－an.若b3＝－2,b10＝12

Answer 1

3=-2,b10=12

7d=b10-b3=14

d=2

bn=-2+2(n-3)=2n-8

a8

=(a8-a7)+(a7-a6)+.+(a2-a1)+a1

=b7+b6+.+b1+3

=[6+4+2+0+(-2)+(-4)+(-6)]+3

=3