Question

等差数列问题 等差数列共2n+1项,奇数项之和132,偶数项之和120,求n

Answer 1

ak=a1+(k-1)d
奇数项
a(2n-1)=a1+(2n-2)d
=a1+(n-1)(2d)
为首项为a1,公差为2d的等差数列,共n+1项
所以
奇S(n+1)=(n+1)a1+n(n+1)(2d)/2
=(n+1)a1+n(n+1)d
=132
偶数项
a(2n)=a1+(2n-1)d
=a2+(n-1)(2d)
为首项为a2,公差为2d的等差数列,共n项
所以
偶Sn=na2+n(n-1)(2d)/2
=na2+n(n-1)d
=na1+n^2*d
=120
(n+1)a1+n(n+1)d=132
na1+n^2*d=120
a1+nd=132/(n+1)
a1+nd=120/n
132/(n+1)=120/n
n=10