Третий и седьмой члены арифметической прогрессии равны 1,1 и 2,3. найти сумму первых десяти её членов.
А₃ = а₁ +2d 1.1 = a₁ + 2d a₁ = 1.1-2d a₁ = 1.1-2d a₇ = a₁ + 6d 2.3 = a₁ +6d 2.3 = 1.1-2d +6d 2.3-1.1 = 4d ₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋₋ a₁ = 1.1-2d a₁ = 1.1-2d a₁ = 1.1-2*0.3 = 0.5 1.2 = 4d d = 0.3 d =0.3 S₁₀ = ((2a₁ + d(n-1))*n)/ 2 = (2*0.5 + 0.3(10-1)) *10) /2 = (1+2.7*10)/2 = = (1+27)/2 = 28/2 = 14 S₁₀ = 14