654. Найдите сумму первых пяти членов геометрической прогрессии (хn), если: 1 a) x₅ = 1, q=; 9 3 б) x₄ = 121,5, q = -3.

a) Дано: x₅ = 1 1/9 = 10/9, q = 1/3 Найти: S₅ Решение: x₅ = x₁ * q⁴ => x₁ = x₅ / q⁴ = (10/9) / (1/3)⁴ = (10/9) / (1/81) = (10/9) * 81 = 10 * 9 = 90 S₅ = x₁ * (1 - q⁵) / (1 - q) = 90 * (1 - (1/3)⁵) / (1 - 1/3) = 90 * (1 - 1/243) / (2/3) = 90 * (242/243) * (3/2) = 45 * (242/81) = 45 * (242/81) = (5 * 242) / 9 = 1210 / 9 = 134.44 S₅ = 90 * (1 - (1/3)⁵) / (1 - 1/3) = 90 * (1 - 1/243) / (2/3) = 90 * (242/243) * 3/2 = 45 * 242 / 81 = 1210 / 9 ≈ 134,44 S5 = a1(1-q^n)/(1-q) a5 = a1q^4 a1 = a5/(q^4) = (10/9)/((1/3)^4) = (10/9)/(1/81) = 90 S5 = 90(1-(1/3)^5)/(1-1/3) = 90(1-1/243)/(2/3) = (270/2)(242/243) = 135(242/243) = 1(24/81) Ответ: S₅ = 1 24/81

б) Дано: x₄ = 121.5, q = -3 Найти: S₅ Решение: x₄ = x₁ * q³ => x₁ = x₄ / q³ = 121.5 / (-3)³ = 121.5 / -27 = -4.5 S₅ = x₁ * (1 - q⁵) / (1 - q) = -4.5 * (1 - (-3)⁵) / (1 - (-3)) = -4.5 * (1 - (-243)) / (1 + 3) = -4.5 * (1 + 243) / 4 = -4.5 * 244 / 4 = -4.5 * 61 = -274.5 \(\frac{243}{4}\) = 60,75\(-4,5\) = -274,5 S5 = x1(1-q^n)/(1-q) x4 = x1q^3 x1 = x4/(q^3) = 121.5/(-3)^3 = 121.5/-27 = -4.5 S5 = -4.5(1-(-3)^5)/(1-(-3)) = -4.5(1+243)/4 = -(9/2)*(244/4) = (9/2)*(61) = 91,125 Ответ: S₅ = 91,125