А7+а9=12 а6*а10=-28 Найти а1 и d

а7=а1+6d a9=a1+8d a6=a1+5d a10=a1+9dсоставим 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