Question

в арифметической прогрессии S4=12,S8=40. Найдите a1 и d

Answer 1

$S_n=frac{2a_1+(n-1)d}{2}*n\\ left { {{S_4=frac{2a_1+3d}{2}*4=12} atop {S_8=frac{2a_1+7d}{2}*8=40}} right. \\ left { {{2a_1+3d=6} atop {2a_1+7d=10}} right. \\ left { {{7d-3d=10-6} atop {2a_1+3d=6}} right. \\ left { {{4d=4} atop {a_1=frac{6-3d}{2}}} right. \\ left { {{d=1} atop {a_1=frac{6-3}{2}}} right. \\ left { {{d=1} atop {a_1=frac{3}{2}}} right.$