Question

数列 (13 11:52:15)

Answer 1

am,a(m+2),a(m+1)成等差数列

即

2a(m+2)=am+a(m+1)

两边同除以a1

2q^(m+1) =q^(m-1) +q^m .(1)

若Sm,S(m+2),S(m+1)成等差数列

2S(m+2)=Sm+S(m+1)

两边同除以a1/(1-q)

2[1-q^(m+2)] =1-q^m +1-q^(m+1)

2q^(m+2)] =q^m +q^(m+1)

两边同除以q

2q^(m+1) =q^(m-1) +q^m 即(1)

===>Sm,S(m+2),S(m+1)成等差数列