Сумма третьего и пятого члена арифметической прогрессии равна 16, а произведение шестого...

$a_1+a_5=16$
$a_1*a_6=64$
$a_1-$ ?
$d-$ ?

$\left \{ {{a_3+a_5=16} \atop {a_1*a_6=64}} \right.$
$\left \{ {{a_1+2d+a_1+4d=16} \atop {a_1*(a_1+5d)=64}} \right.$
$\left \{ {{2a_1+6d=16} \atop {a^2_1+5a_1*d=64}} \right.$
$\left \{ {{a_1+3d=8} \atop {a^2_1+5a_1*d=64}} \right.$
$\left \{ {{a_1=8-3d} \atop {(8-3d)^2+5d(8-3d)=64}} \right.$
$\left \{ {{a_1=8-3d} \atop {64-48d+9d^2+40d-15d^2=64}} \right.$
$\left \{ {{a_1=8-3d} \atop {-8d-6d^2=0}} \right.$
$\left \{ {{a_1=8-3d} \atop {4d+3d^2=0}} \right.$
$4d+3d^2=0$
$d(4+3d)=0$
$d=0$ или $d=- \frac{4}{3}$

$\left \{ {{d=0} \atop {a_1=8-3*0}} \right.$ или $\left \{ {{d=- \frac{4}{3} } \atop {a_1=8-3*(- \frac{4}{3}) }} \right.$
$\left \{ {{d=0} \atop {a_1=8}} \right.$ или $\left \{ {{d=- 1\frac{1}{3} } \atop {a_1=12 }} \right.$