公差不为0的等差数列{an｝中,已知前n项的和为Sn,若S8=S5+45,且a4,a7,a12成等比数列 求数列{an}

S8=S5+a6+a7+a8
因为a6+a8=2a7
所以,S8=S5+3a7
已知S8=S5+45
所以,a7=15
又已知a4,a7,a12成等比数列,则：a4*a12=a7^2=225
===> (a7-3d)*(a7+5d)=225
===> (15-3d)*(15+5d)=225
===> 225+30d-15d^2=225
===> 15d^2-30d=0
===> d=2【已知d≠0】
所以,a7=a1+6d=15
那么,a1=3
所以,等差数列的通项为：an=a1+(n-1)d=3+(n-1)*2=2n+1.
所以,Sn=na1+n(n-1)d/2=3n+n^2-n=n^2+2n
则,bn=1/Sn=1/(n^2+2n)=1/[n*(n+2)]=(1/2)*[(1/n)-(1/n+2)]
所以,Tn=(1/2)*[1-(1/3)+(1/2)-(1/4)+(1/3)-(1/5)+……+(1/n)-(1/n+2)]
=(1/2)*[1+(1/2)-(1/n+1)-(1/n+2)]
=(3n^2+7n+3)/(4n^2+12n+8)

解 S8=S5+45
S8-S5=45
a6+a7+a8=45
因为在公差不为0的等差数列{an｝中
所以a6+a8=2a7
3a7=45
a7=15
a4，a7，a12成等比数列
所以有a7^2=a4*a12
a4*a12=22...