Question

在等差数列{an}中,已知a1+a2=12,a4=7,求a9

Answer 1

a2=a1+d a4=a1+3d=7
ai+a2=a1+a1+d=2a1+d=12
可得a1=5.8
d=o.4
a9=a1+8d=9

Answer 2

a1+a2=12
a1+a1+d=12
2a1+d=12
a4=a1+3d=7
d=2/5
a1=29/5
a9=29/5+8*2/5=9

Answer 3

因为等差数列{an}中a2=6,a5=0
所以3d=-6
d=-2
所以a1=8,a3=4,a6=-2,a9=-8
因为a3 a6 a9......成公差为-6的等差数列
所以cn=8 (n=1)
16-6n (n>1)
因为a1=a2-d=8,
n=1时为特例c1=a1=8

Answer 4

a1+a2=12
a1+d=a2所以
2*a1+d=12
a1+3*d=a4=7所以d=0.4,a1=5.8
a9=a1+8*d=5.8+8*0.4=9