Question

в геометрической прогрессии   ( A n )  a 10=27, а12=108. найдите а11.

Answer 1

$<var>a_{10}=27 \\ \\ a_{12}=108 \\ \\ a_{11}^2=a_{10}*a_{12} \\ \\ a_{11}^2=27*108 \\ \\ a_{11}=б\sqrt{27*108}=б\sqrt{3*3*3*3*36}=б9*6=б54</var>$

 

Ответ: -54; 54