1.解:令首项为x,公差为d,
则此数列为:x,x+d,x+2d,...,x+(2n-1)d
令奇数项之和为A=x+【x+2d】+...+[x+(2n-2)d]=24
偶数项之和为B=[x+d]+[x+3d]+...+[x+(2n-1)d]=30
B-A=d+d+,,,+d=nd=6
又∵[x+(2n-1)d]-x=2nd-d=10.5
∴ d=1.5n=4
A+B={x+[x+(2n-1)d]}/2*2n=54 得x=1.5
综上可得:此数列的首项为1.5,公差为1.5,项数为8.2.解:因为{an},{bn}都为等差数列
所以令n=19
S19=(a1+a19)/2*19=a10*19
M19=(b1+b19)/2*19=b10*19
S19/M19=a10/b10=(3*19+2)/(2*19-1)=59/37 |