已知正数a1,a2,a3.a7构成等比数列,若前5项的和为7倍根号2加6,后五项的和为14倍根号2加12,则a6?

答：设公比为q,则：a1+a2+a3+a4+a5=a1(1+q+q^2+q^3+q^4)=7√2+6a3+a4+a5+a6+a7=a1q^2(1+q+q^2+q^3+q^4)=14√2+12所以q^2=2,因为an为正数,所以q>0,即q=√2.a1(1+q+q^2+q^3+q^4)=a1(√2^5-1)/(√2-1)=7√2+6所以a1=(8-...