Question

{an}是等差数列 , 且a1+a2+a3=35 , a2+a3+a4=63 , 求a3+a4+a5=?

Answer 1

a2-a1=a3-a2=a4-a3=d
所以(a2+a3+a4)-(a1+a2+a3)=3d=28
同理
a3+a4+a5=(a2+a3+a4)+3d=91

Answer 2

an= a1+(n-1)d
a1+a2+a3=35
3a1+3d=35 (1)
a2+a3+a4=63
3a1+6d=63 (2)
(2)-(1)
3d=28
d=28/3
a1=7/3
a3+a4+a5
=3a1+9d
=3(7/3)+9(28/3)
=7+84
=91