3)дана арифметическая прогрессия , a5=8,7, a8=12,3. Найти d и a1

d=1,2

a1=3,9
$a_{n} =a_{1} + d (n - 1)$
a2=a1+1d
a3=a1+d2
a4=a1+3d
a5=a1+4d
a6=a1+5d
a7=a1+6d
a8 = a1 + 7d

$\left \{ {a8=a1+7d} \atop {a5=a1+4d}} \right.$

$\left \{ {{12,3=a1+7d} \atop {8,7=a1+4d}} \right.$

$\left \{ {{a1=12,3-7d} \atop {8,7=12,3-7d+4d}} \right.$

8,7=12,3-3d
3d=12,3-8,7
d=1,2

a1=12,3-6*1,2
a1=3,9