設自然序列為:A1，A2，…A(2n+1)，
那麽2n+1數的中項=a /(2n+1)
b = {[a(n+2)]^2-[a1]^2}+…+{[a(2n+1)]^2+[ an]^2}
=(n+1)*[A 1+A(n+2)]+…+(n+1)*[An+A(2n+1)]
=(n+1)[A 1+A2+…+An+A(n+2)+…+A(2n+1)]
=(n+1)[a - a/(2n+1)]
=2na(n+1)/(2n+1)
所以，因為b/a=60/11，2n(n+1)/(2n+1)= 60/11。
簡化:11n 2-49n-30 = 0
(n-5)*(11n+6)=0
解法:n = 5；(負根去掉)
答案:n=5