Question

геометриялык прогрессия

Answer 1

$b_{11}=b_5*q^{(11-5)}; \ 1647=61*q^6; \ q^6=1647:61=27=3^3; \ q=3^{ frac{3}{6}} =3^{ frac{1}{2}} = sqrt{3}$

$b_{11}=b_9*q^{(11-9)}=b_9*q^2; \ \ b_9= frac{b_{11}}{q^2} = frac{1647}{( sqrt{3} )^2} =frac{1647}{3} =549 ;$

$b_5=b_2*q^{(5-2)}=b_2*q^3; \ \ b_2= frac{b_5}{q^3}= frac{61}{( sqrt{3} )^3}= frac{61}{3 sqrt{3} }$.