(1)设等差数列{a n}的公差为d,由a 7+a 9=16,a 4=1,可得
2a 1 +14d=16
a 1 +3d=1 ,解得
d=
7
4
a 1 =-
17
4 .
∴ a 12 =-
17
4 +11×
7
4 =15.
(2)S 15= 15×(-
17
4 )+
15×14
2 ×
7
4 =120.
(1)设等差数列{a n}的公差为d,由a 7+a 9=16,a 4=1,可得
2a 1 +14d=16
a 1 +3d=1 ,解得
d=
7
4
a 1 =-
17
4 .
∴ a 12 =-
17
4 +11×
7
4 =15.
(2)S 15= 15×(-
17
4 )+
15×14
2 ×
7
4 =120.