1.若两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且满足Sn/Tn=(7n+3)/(n+3),则a8/b8=


2.设{an}是正数组成的数列,其前n项和为Sn,.并且对于所有的n 属于N+,an与2的等差中项等于Sn与2的等比中项.
1) 写出数列{an}的前3项

2) 求数列{an}的通项公式(写出推证过程)

a1+a15 =a2+a14=....=a7+a9 =2a8
==>S15 = 15a8
T15 =15b8
a8/b8=S15/T15 =(7*15+3)/(15+3) =108/18=6


2
(a1+2)/2 =根号（2S1） =根号（2a1）
==>(a1+2)^2 =8a1
an是正数组成的数列==>a1=2

(a2+2)/2 =根号（2S2）=根号[2(a2+ 2)]
==>(a2+2)^2 =8(a2+2)
==>(a2+2)(a2+2 -8)=0
an是正数组成的数列 ==>a2 =6

(a3+2)/2 =根号（2S3）=根号[2(a2+ 8)]
==>(a3+2)^2 =8(a3+8)
an是正数组成的数列 ==>a3=10


2) 求数列{an}的通项公式
(an+2)/2 =根号（2Sn）
===>Sn =(an+2)^2/8 ....(1)
S(n-1) =[a(n-1)+2)^2/8 ...(2)
(1)-(2)==>
an = {(an+2)^2 -[a(n-1)+2)^2]}/8
(an)^2 -[a(n-1)]^2 =4an+4a(n-1)
[an +a(n+1)][an -a(n+1)]=4[an +a(n+1)]
{an}是正数组成的数列
==>an +a(n+1)不等于0，两边除之
==> an -a(n+1) =4
{an}是公差为4的等差数列
a1 =2
==>通项公式an =2+(n-1)*4 =4n-2