Варифметической прогрессии an a2 + a7 =14, s10 =80. найти d

A2+a7=14, s10=80, d=?
a2=a1+d, a7=a1+6d
a2+a7=a1+d+a1+6d=2a1+7d
2a1+7d=14
7d=14-2a1
s10=10(a1+a10)/2
s10=5(a1+a10)
80=5(a1+a1+9d)
2a1+9d=16

7d=14-2a1
9d=16-2a1

9d-7d=16-2a1-(14-2a1)
2d=2, d=1
Otvet: d=1

(a1=(14-7d)/2=(14-7)/2=3,5
a7=3,5+6=9,5, a2=3,5+1=4,5, a2+a7=4,5+9,5=14
a10=3,5+9=12,5,s10=10(3,5+12,5)/2=5(3,5+12,5)=80)

А1+d+a1+6d=14
2a1+7d=14

2a1+9d
S10 = • 10 = (2a1+9d)•5 = 80
2

2a1+9d=16
2a1+7d=14
2d=2
d=1