Question

已知等差数列{an}中,s20=120,求a3+a18

Answer 1

S(20)=(a1+a20)*20/2=10(a1+a20)=10(a3+a18)

10(a3+a18)=120

a3+a18=12

对等差数列{an}

有若i+j=m+n

则a(i)+a(j)=a1+(i-1)d+a1+(j-1)d

=2a1+(i+j-2)d

=2a1+(m+n-2)d

=a1+(m-1)d+a1+(n-1)d

=a(m)+a(n)

即下标和相等的任意两项相加之和都相等