{an}为等差数列,设公差为d,有d=a2-a1=a3-a2.则2a2=a3+a1
又a1=x,a2=2x,a3=5x-4,有2*2x=(5x-4)+x,得x=2.
故a1=2,a2=4,a3=6,d=a2-a1=2.
由{an}为等差数列有Sn=a1n+[n(n-1)d/2],则由Sk=a1k+[k(k-1)d/2]=2k+[k(k-1)*2/2]=k^2+k=2550,得k=50.