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*a41.求a1与d2.求数列｛An｝的前n项和Sn

1.因为｛An｝为等差数列
所以a3=a2+a4*½
a2+a3+a4=15=3a3 a3=5
(5-1)^2=16=(5-d)(5+d)
d=3 (-3舍去)
a1=5-2d=-1 d=3
2.
an=a1+(n-1)*d=3n-4
Sn=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

解
1、由a2+a3+a4=15 与2a3=a2+a4得a3=5，
又（a3-1）的平方=a2*a4
故a2*a4=16，再结合a2+a4=2a3=10得a2=2，a4=8
所以d=3，a1=-1
2、
Sn=na1+n(n-1)d/2=-n+n(n-1)3/2=3/2*n^2-5/2*n

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...