Question

在等差数列{an}中，已知公差d=1/2，且a1+a3+a5+…+a99=60，则a1+a2+a3+…+a99+a100=？

Answer 1

a1+a3+a5+…+a99=60，
a2+a4+a6+...+a100=(a1+1/2)+(a3+1/2)+...+(a99+1/2)=(a1+a3+a5+…+a99)+50x1/2=60+25
=85
所以，a1+a2+a3+…+a99+a100=60+85=145

Answer 2

a2=a1+1/2,a4=a3+1/2,...a100=a99+1/2.....
所以a1+a2+...+a100=(a1+a3+...+a99)*2+50*1/2=145

Answer 3

a2+a4+a6……+a100
=(a1+d)+(a3+d)+(a5+d)+……+(a99+d)
=(a1+a3+a5+…+a99)+50d
=60+50×1/2
=85
a1+a2+a3+…+a99+a100
=(a1+a3+a5+…+a99)+ (a2+a4+a6……+a100)
=60+85
=145

Answer 4

a1+a3+a5+…+a99=60
a2+a4+a6+…+a100=(a1+d)+(a3+d)+(a5+d)+…+(a99+d)=60+ 1/2 * 50 =85
a1+a2+a3+a4+…+a100=a1+a3+a5+…+a99+a2+a4+a6+…+a100=60+85=145