Question

设等差数列{an}共有2n+1项，所有奇数之和为132，所有偶数项之和为120，则n=

请用种简单的方法

Answer 1

an = a1+(n-1)d
a1+a3+...+a(2n+1) = 132
(a1+nd)(n+1)=132 (1)

a2+a4+.....+a(2n)=120
(a1+nd)n =120 (2)
(1)/(2)
(n+1)/n = 11/10
10n+10=11n
n =10