（1）数列{an}是公比不为1的等比数列且a5,a3,a4成等差数列,则2a3=a4+a5,即q^2+q-2=0,解得q=1(舍）或q=-2
（2）S（k+2)+S(k+1)=[a1(1-q^(k+2)]/(1-q)+[a1(1-q^(k+1)]/(1-q)=[a1/(1-q)][2-q^(k+2)-q^(k+1)] =[a1/(1-q)][2-q^k(q^2+q)]=[a1/(1-q)][2-2q^k]
Sk=[a1/(1-q)][1-q^k
故S（k+2)+S(k+1)= 2Sk即对任意k(正数）,s(k+2),sk,s(k+1)成等差数列