等差数列{an}中,a5=0.3,a12=3.1,求a18+a19+a20+a21+a22的值
问题描述:
等差数列{an}中,a5=0.3,a12=3.1,求a18+a19+a20+a21+a22的值
有助于回答者给出准确的答案
答
A5=A1+4D=0.3
A12=A1+11D=3.1
7D=2.8 D=0.4
A18+A19+A20+A21+A22=5A1+(17+18+19+20+21)D
=5A1+95D
=5A1+55D+40D
=5A12+40D
=15.5+40*0.4
=15.5+16
=31.5