设{An}为等差数列,公差d为正数,已知a2+a3+a4=15 又(a3-1)的平方=a2*a4

问题描述:

设{An}为等差数列,公差d为正数,已知a2+a3+a4=15 又(a3-1)的平方=a2*a4
1.求a1与d
2.求数列{An}的前n项和Sn

1.{An}为等差数列,所以2a3=a2+a4a2+a3+a4=15=3a3 a3=5 (5-1)^2=16=(5-d)(5+d)d=3 (-3舍掉)所以a1=5-2d=-1 d=32.an=a1+(n-1)*d=3n-4Sn=3-4+6-4+9-4+...+3n-4=3*(1+2+3+...n)-4n=3*n*(n+1)/2-4n=3/2*n^2-5/2*n...