Question

在项数为2n+1的等差数列中，所有奇数项之和为165，所有偶数项之和为150则n的值为？

Answer 1

a(k)=a+(k-1)d,
a(2k)=a+(2k-1)d=a+(k-1)(2d)+d=(a+d)+(k-1)*(2d),
150=a(2)+a(4)+...+a(2n)=(a+d)n+(n-1)n*(2d)/2=(a+d)n+(n-1)nd=n[a+d+(n-1)d]=n[a+nd].
a(2k-1)=a+(2k-2)d=a+(k-1)*(2d)
165=a(1)+a(3)+...+a(2n+1)=(n+1)a+n(n+1)*(2d)/2=(n+1)a+n(n+1)d=(n+1)[a+nd].
150/165=n[a+nd]/[(n+1)(a+nd)]=n/(n+1),
150(n+1)=165n,
150=15n,
n=10