等差数列{an}中,sn表示前n项之和,s10=s8,a1=17/2,求a1~an各项绝对值的和是多少

Question

等差数列{an}中,sn表示前n项之和,s10=s8,a1=17/2,求a1~an各项绝对值的和是多少

Answer 1

s10=s8+a9+a10 = s8
故a9=-a10=-(a9+d)
也即2*a9 = -d
a9 = a1+8d = 17/2 + 8*d
联立解得d=-1
an = a1+(n-1)d = 17/2 + 1 -n = 19/2-n
当n0
sn = [17/2 + 17/2 + (n-1)*(-1)]*n/2 = (18-n)*n/2
=-n*n/2 + 9n
s9= (18-9)*9/2 = 81/2
当n>9的时候,an