Question

若數(shù)列{a_n}滿足

1

an+1

-

1

an

=d（n∈N^*，d為常數(shù)），則稱數(shù)列{a_n}為調(diào)和數(shù)列．記數(shù)列{

1

xn

}為調(diào)和數(shù)列，且x₁+x₂+…+x₂₀=200，則x₅+x₁₆=

 

．

Answer 1

分析：由題意知道，本題是構(gòu)造新等差數(shù)列的問(wèn)題，經(jīng)過(guò)推導(dǎo)可知{x_n}是等差數(shù)列，運(yùn)用等差數(shù)列的性質(zhì)可求解答案．

解答：解：由題意知：
∵數(shù)列{

1

xn

}為調(diào)和數(shù)列
∴

1

1 xn+1

-

1

1 xn

=xn+1-xn=d
∴{x_n}是等差數(shù)列
 又∵x₁+x₂+…+x₂₀=200=

20(x1+x20)

2

∴x₁+x₂₀=20
又∵x₁+x₂₀=x₅+x₁₆
∴x₅+x₁₆=20
故答案為20．

點(diǎn)評(píng)：本題主要考查新數(shù)列定義，及等差數(shù)列的重要性質(zhì)，屬中檔題型．