Question

等差数列{AN}中，公差为1/2，且A1+A3+A5······+A99=60，则A1+A2+A3······A99+A100等于多少？

要有过程。

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

因为公差为1/2，所以
A2-A1=1/2
A4-A3=1/2
A6-A5=1/2
……
A100-A99=1/2
(A2+A4+…+A100)-(A1+A3+…+A99)=99×1/2=99/2
所以
A2+A4+…+A100=99/2+60=109.5
A1+A2+A3+…+A99+A100=109.5+60=169.5

Answer 3

请问A1是多少？你给的条件不足以构成等差数列......

Answer 4

A2=A1+1/2
A4=A3+1/2
A6=A5+1/2
……
A100=A99+1/2

A2+A4+A6+……+A100=(A1+1/2)+(A3+1/2)+(A5+1/2)+……+(A99+1/2)
=(A1+A3+A5······+A99)+50×1/2
=60+25
=85

A1+A2+A3+······+A99+A100
=(A1+A3+A5······+A99)+(A2+A4+A6+……+A100)
=60+85
=145