已知等差数列{an}的前n项和为Sn=pn²-2n+q（p,q∈R,n∈N*）

1)A1=S1=p-2+q,
An=Sn-Sn-1=pn^2-2n+q-p(n-1)^2+2(n-1)-q=p(2n-1)-2=2pn-p-2
d=An-An-1=2p
An=A1+(n-1)*d=p-2+q+2(n-1)p=2pn-p-2+q = 2pn-p-2
q=0
Sn=pn^2-2n,An=2pn-p-2,d=2p,A1=p-2
2) A3=A1+2d=p-2+4p=5p-2=18,p=4
A1=2,An=8n-6,
8n-6=2log2Bn,log2Bn=4n-3,Bn=2^(4n-3),Bn/Bn-1=2^4=16,Bn是公比为16的等比数列,首项B1＝2,设Bn的前n项和是Tn
Tn=2*(16^n-1)/(16-1)=2*(16^n-1)/15

答案如下图：
http://dl.zhishi.sina.com.cn/upload/11/60/93/1566116093.14073902.JPG

an=sn-s(n-1)=p(2n-1)-2,
a1与a5的等差中项
a3=5p-2=18,p=4,
an=8n-6,8n-6=2log2 * bn,
bn=2^(4n-3),b(n+1)/bn=2^4,
{bn}是等比数列,b1=2,q=2^4,Tn=2^(4n-1)/15