1.
an=a1+(n-1)d
a15=a1+14d=9
a1=9-14d
sn=na1+n(n-1)d/2
=n(9-14d)+n(n-1)d/2
=(1/2)dn^2+(9-29d/2)n
=(1/2)d[n+(9-29d/2)/d]^2-(1/2)d[(9-29d/2)/d]^2
=(1/2)d[n+(9-29d/2)/d]^2-(1/2)(9-29d/2)^2/d
当d＜0,n+(9-29d/2)/d≈0时,才有最大值,
n=-(9-29d/2)/d
16＜-(9-29d/2)/d＜18
-16d＜9-29d/2＜-18d
-32d-18＜-29d＜-36d-18
36d+18＜29d＜32d+18
-6＜d＜-18/7;
2.
-6＜d＜-18/7
-5≤d≤-3
(1)d1=-5,a1=79,an=79-5(n-1)=84-5n
……
(2)d2=-4,a1=65,an=65-4(n-1)=69-4n
……
(3)d3=-3,a1=51,an=51-3(n-1)=54-3n
……