Question

已知{an}为等差数列,且a1＋a3＝8,a2＋a4＝12 (1)求an的通项公式

Answer 1

a2+a4-(a1+a3)＝12-8 2d=4 d=2

a1+a1+2d=8 a1=2

an=a1+(n-1)d=2n

a1,ak,Sk＋2等比

Sn=na1+n(n-1)d/2=n²+n

ak²=a1*S(k+2)

(2k)²=2*[(k+2)²+k+2]

k=-1(舍去),k=6