已知数列an是等比数列,其中a7=1,且a4,a5+1,a6成等差数列

a7=aq^6=1aq^4=1/q^2aq^3=1/q^3aq^5=1/qa4,a5+1,a6成等差数列2(a*q^4+1)=a*q^3+a*q^52a*q^4+2=a*q^3+a*q^52/q^2+2=1/q^3+1/q2q+2q^3=1+q^2q^2(2q-1)+(2q-1)=0(q^2+1)(2q-1)=0因为q^2+1不等于0所以2q-1=0q=1/2aq^6=1a...