Question

在等差数列{an}中,公差为1/2,且a1+a3+a5+...+a99=60,则a1+a2+a4+...+a100等于?

Answer 1

a1+a3+a5+···+a99
=(a1+a99)*50/2=60
a1+a99=2.4
a1+a100=a1+(a99+d)=2.4+1/2=2.9

a1+a2+a3+···+a100
=(a1+a100)*100/2
=145

Answer 2

解:an=a1+1/2(n-1)
a1+a3+a5+....+a99=60
50(a1+a99)/2
=50[a1+a1+1/2(99-1)]/2
=50(2a1+49)/2
=60
a1= - 233/10
=-23.3
a1+a2+a4+.....+a100
=a1+50(a2+a100)/2
=a1+50[(a1+1/2)+a1+99/2]/2
=a1+50(2a1+50)/2
= - 233/10+50[2*(-233/10)+50]/2
=617/10
=61.7
正确答案如上，欢迎采纳！感谢了！