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

d=1,2 a1=3,9 [latex] a_{n} =a_{1} + d (n - 1)[/latex] a2=a1+1d a3=a1+d2 a4=a1+3d a5=a1+4d a6=a1+5d a7=a1+6d a8 = a1 + 7d [latex] \left \{ {a8=a1+7d} \atop {a5=a1+4d}} \right. [/latex] [latex] \left \{ {{12,3=a1+7d} \atop {8,7=a1+4d}} \right. [/latex] [latex] \left \{ {{a1=12,3-7d} \atop {8,7=12,3-7d+4d}} \right. [/latex] 8,7=12,3-3d 3d=12,3-8,7 d=1,2 a1=12,3-6*1,2 a1=3,9