等差数列{an}.a1+a7=22.a4+a10=40求公差d

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等差数列{an}.a1+a7=22.a4+a10=40求公差d
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等差数列{an}.a1+a7=22.a4+a10=40求公差d
等差数列{an}.a1+a7=22.a4+a10=40求公差d

等差数列{an}.a1+a7=22.a4+a10=40求公差d
a1+a7=2a4=22,则:a4=11.a4+a10=40,得:a10=29,a10-a4=6d=18,得:d=3

2a1+6d=22
2a1+12=40
6d=18
d=3

a4=11
a7=20
11+3d=20
d=3

求:1.等差数列{an} 2.若数列{bn}是等差数列,bn=Sn/(n+c),求非零a2+a5=22=a3+a4 a3*a4=117 所以解得a3=9,a4=13 所以公差d=a

a7=a1+6*d
a4=a1+3*d
a10=a1+9*d
代入
a1+a7=22,a4+a10=40
a1+a1+6*d=22 (1)
a1+3*d+a1+9*d=40 (2)
(2)式-(1)式
3*d+9*d-6*d=40-22
6*d=18
d=3

设an=a1+(n-1)d,则a1+(a1+6d)=22
(a1+3d)+(a1+9d)=40
两式相减可解得d=3

(a4+a10)-(a1+a7)=18= a4-a1+a10-a7= 3d+3d,公差d为3.

a1+3d=a4
a7+3d=a10
a1+a7+6d=a4+a10
6d=18
d=3