Question

Номер 6-7 срочно!!!!!!"!

Answer 1

$6.\;\;a_1=3,\;a_{n+1}=a_n+4\Rightarrow d=4\\a_10=a_1+9d=3+9\cdot4=3+36=39\\7.\;\;(a-3)^2-a(5a-6)=a^2-6a+9-5a^2+6a=-4a^2+9\\a=-\frac12\Rightarrow-4a^2+9=-4\left(-\frac12\right)^2+9=-4\cdot\frac14+9=-1+9=8$