等差数列an中 d=1 a1+a2+a3+…+a100=100 求 a4+a8+a12+…a100的值_数学解答

等差数列an中 d=1 a1+a2+a3+…+a100=100 求 a4+a8+a12+…a100的值

人气：273 ℃　时间：2020-08-25 08:52:31

解答

a1+a2+a3+…+a100
=100a1+99*100/2d
=100a1+99*50=100
a1=-97/2
a4=a1+3d=-97/2+3=-91/2
a100=a1+99d=-97/2+99=101/2
a4+a8+a12+…a100
=(-91/2+101/2)*25/2
=125/2