Question

高中数学，求解，要过程，谢谢啦～

Answer 1

Answer 2

由题意知等差数列公差d≠0；等差数列a1+a2+a3+a4+a5+a6=60可得出6a1+15d=60即2a1+5d=20;又a6²=a1*a21得出（a1+5d）²=a1*(a1+20d)化简为a1=5/2d即可得出a1=5,d=2.数列an=a1+(n-1)d=5+(n-1)*2=2n+3;Sn=5n+n(n-1)=n(n+4)
bn+1-bn=an;bn-bn-1=an-1;bn-1-bn-2=an-2......b2-b1=a1等式左边全加起来便有bn+1-b1=an+....a1=Sn=n(n+1)则得出bn通项公式bn=Sn-1+3即1/bn=1/((Sn-1)+3)=1/(n²+2n+3-3)=1/n -1/(n+2)则数列（1/bn）的前n项和Tn=1/n-1/(n+2)+1/(n-1)-1/(n+1)+1/(n-2)-1/n+1/(n-3)-1/(n-1)......+1/4-1/6+1/3-1/5+1/2-1/4相互抵消=1/2+1/3-1/（n+1）-1/(n+2)剩下自己化简吧！

Answer 3