1、 解a1+a3+a5+···+a(2k-1)+a(2k+1)=216a2+a4+a6+···+a(2k) =192
a1+kd+a2+a4+a6+···+a(2k)=216,a(k+1)=24
S(2k+1)=408=(a1+a(2k+1))(2k+1)/2=a(k+1)*(2k+1)=24*(2k+1),2k+1=17
k=8,a1=1,所以d=23/8,an=1+(23/8)(n-1),a17=47
2、依題意 得1/3 S3+1/4 S4 =2,2a1+a3+a4=4,
1/3 S3*1/4 S4 =1/5S5 * 1/5S5 即 (a1+a3)(a1+a4)=(a1+a5)(a1+a5)
(a1+d)(2a1+3d )=2(a1+2d)(a1+2d),d=0,d=-a1
d=0時(shí),a1=1
d=-a1時(shí) a1=-2(舍棄),所求數(shù)列的通項(xiàng)公式為an=1