Question

几道数学题(有关等差数列)

1.等差数列{an},S20=170,则a6+a9+a11+a16=
2.等差数列{an},S9=18,Sn=240,且a(n-4)=30,求n(n≥5)
3.等差数列{an},d=1,S99=99,求a3+a6+a9+……+a99的值.

Answer 1

1.a6+a9+a11+a16=2(a10+a11)=2(a1+a20)=34

2. a1+a9=2a5=4
a1+an=480/n
a5+a(n-4)=480/n
a5=2 a(n-4)=30 带入得n=15

3.s99=99 => a1+a99=2 =>a50=1 =>2a51=4
a3+a6+a9+...+a99=15*(2*a56)=60

Answer 2

S20=10(a1+a20)=10(a10+a11)=170
a1+a20=a10+a11=17
a6+a9+a11+a16= 2(a1+a20)=34

S9=(a1+a9)+(a2+a8)+…+(a4+a6)+a5
=9a5
a5=2
a1+an=a5+a(n-4)=32
Sn=(a1+an)*d/2
d=2Sn/(a1+an)=15