Question

高一数学题求解!!!!!!!

1.在等差数列{an}中.已知a3+a99=200,求S101. 2.在等差数列{an}中a15+a12+a9+a6=20.求S20 3.若数列{an}为等比数列.a1+a2+a3=7.a1a2a3=8.求an

Answer 1

a1+a101=a3+a99=200 S101=101×（a1+a101)除2=10100
a15+a12+a9+a6=2(a1+a20) S20=20×（a1+a20)÷2=100

Answer 2

1.s101=(a3+a99)*101/2=10100
2.s20=(a15+a12+a9+a6)*20/4=100
3.an=2^(n-1)

Answer 3

1 .10100 2.100 3.an=2的n-1 次方

Answer 4

1，10100
2，100
3，2的n-1次

Answer 5

S101=101a51=101(a3+a99)/2=10100
s20=10(a1+a20)=5(a15+a6+a12+a9)=100
a1=1 a2=2 a3=4 or a1=4 a2=2 a3=1
an=2^(n-1) or an=2^(3-n)