А7+а9=12

а6*а10=-28

Найти а1 и d

а7=а1+6d

a9=a1+8d

a6=a1+5d

a10=a1+9d

составим систему:

a7+a9 = а1+6d + a1+8d=12

a6*a10 = (a1+5d) * (a1+9d) = - 28

2a1+14d=12

a1^2+14a1d+45d^2=-28

a1+7d=6

a1^2+14 d a1+45d^2=-28

a1=6-7d

(6-7d) ^2+14d (6-7d) + 45d^2=-28

a1=6-7d

36-84d+49d^2+84d-98d^2+45d^2=-28

a1=6-7d

36-4d^2=-28

a1=6-7d

9 - d^2=-7

a1=6-7d

d=±4

a1=6-7*4=-22

d=4

a1=6-7 * (-4) = 34

d=-4

А7=а1+6d

a9=a1+8d

a6=a1+5d

a10=a1+9d

a1+6d+a1+8d=12

2a1+14d=12

a1+7d=6

a1=6-7d

(6-7d+5d) (6-7d+9d) = - 28

(6-2d) (6+2d) = - 28

36-4d^2=-28

-4d^2=-64

d^2=16

d=4; d=-4

а1=-22; d=34

При d=4; a1=-22:

а7=а1+6d=-22+24=2

a9=a1+8d=-22+32=10

a6=a1+5d=-22+20=-2

a10=a1+9d=-22+36=14

При d=-4; a1=34:

а7=а1+6d=34-24=10

a9=a1+8d=34-32=2

a6=a1+5d=34-20=14

a10=a1+9d=34-36=-2

Оба значения подходят к решению