Question

已知数列{an}为等差数列，a1+a3+a5=105,a2+a4+a6=99,则a20=?

我要具体过程哦

Answer 1

a1+a3+a5=3a3=105,
a3=35
a2+a4+a6=3a4=99,
a4=33
d=-2,
an=35-2(n-3)=41-2n
a20=41-2*20=1

Answer 2

2式-1式得3d=-6
d=-2
一式可变形为：a1+a1+2d+a1+4d=105
及3a1+6d=105
所以a1=39
所以a20=a1+19d=1

Answer 3

a2-a1+a4-a3+a6-a5=3d=99-105=-6
d=-2

a1+a1+2d+a1+4a=3a1+6d=105

a1=39

a20=a1+19d=1