Question

геометрик прогрессия b3=3 b7=27 найдите b15=?

Answer 1

$Bn=B1*q^{n-1}$
$B3 = B1*q^{3-1}=B1*q^{2}$
$B7 = B1*q^{7-1} = B1*q^{6}$
$frac{B7}{B3}= frac{q^{6}}{q^{2}}=q^{6-2}=q^4$
$frac{27}{9} = q^4 - textgreater q= sqrt[4]{3}$
$B1= frac{B3}{q^2} - textgreater B1= frac{9}{q^2}$
$B15=B1*q^{14}= frac{9}{q^2} *q^{14}=9*q^{12}$
$9*( sqrt[4]{3})^{12} = 9 *3^{12/4}=9*3^{3}=243$
$B15 = 243$