下有数列一题,

S3=a1+a2+a3=a1(1+q+q²)=168 ①a2-a5=a1(q-q^4)=a1q(1-q³)=a1q(1-q)(1+q+q²)=42 ②②÷①,得：q(1-q)=1/4,整理得：4q²-4q+1=0,即(2q-1)²=0,∴q=1/2,代入①式,得：a1*(1+1/2+1/4)=168∴a1=96,∴a6=a1*q^5=96×(1/2)^5=3∴a5和a7的等比中项为±a6,即为±3