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

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