Question

数列问题11111111等差数列{an}的项数为奇数,S奇=60；S偶=45,则其项数为___.

Answer 1

设项数为2k+1,

S奇＝(k+1)[a1+a(2k+1)]/2=60

S偶＝k[a2+a(2k)]/2=k[a1+a(2k+1)]/2=45

两式相除得

(k+1)/k=60/45

k=3

项数为2k+1=7