Question

已知某等差数列共有2n+1项，其奇数项之和为630，偶数项之和为600，则此数列的项数为（　　） A．40 B．41 C

Answer 1

∵奇数项和S ₁=

( a 1 + a 2n+1 ) (n+1)

2 =630；偶数项之和S ₂=

( a 2 + a 2n ) •n

2 =600

∴

S 1

S 2 =

( a 1 + a 2n+1 ) (n+1)

2

( a 2 + a 2n ) •n

2 =

n+1

n =

630

600

∴n=20

∴2n+1=41

故选B．