b1, q ---геометрическая прогрессия
b2 = b1*q, b3 = b1*q^2, b4 = b1*q^3
а1, d ---арифметическая прогрессия
a2 = a1+d, a3 = a1+2d, a4 = a1+3d
a1=b1, a2=b2+4, a3=b3+5, a4=b4
a2=b2+4 => b1+d = b1*q+4
a3=b3+5 => b1+2d = b1*q^2+5
a4=b4 => b1+3d = b1*q^3
система из трех уравнений...
из первого: b1*(q-1) = d-4
из третьего вычтем второе: d = b1*q^2*(q-1)-5 = q^2*(d-4)-5
q^2*(d-4) = d+5 => q^2 = (d+5) / (d-4)
из второго: b1-b1*q^2 = 5-2d
b1*(1-q^2) = 5-2d
b1*(1- (d+5) / (d-4) ) = 5-2d
b1*( 9 / (4-d) ) = 5-2d
b1 = (5-2d)(4-d) / 9
подставим в первое: (5-2d)(4-d) * (q-1) = 9(d-4)
q-1 = 9 / (2d-5)
q = 9 / (2d-5) + 1 = (2d+4) / (2d-5) => q^2 = (2d+4)^2 / (2d-5)^2 =>
(2d+4)^2 / (2d-5)^2 = (d+5) / (d-4)
(2d+4)^2 * (d-4) = (2d-5)^2 * (d+5) отсюда d = 7
q = (2*7+4) / (2*7-5) = 18/9
q = 2
a1 = b1 = (5-2*7)(4-7) / 9 = 3
геометрическая прогрессия: 3, 6, 12, 24
арифметическая прогрессия: 3, 10, 17, 24
6+4 = 10... 12+5 = 17...
1) 6 1/2 - 4 1/4 = 6 2/4 - 4 1/4 = 2 1/4
2) 2 1/4 : 2 1/2 = 9/4 : 5/2 = 9/4 * 2/5 = 9/10
(4 8/15 - 1 1/3) * 1 7/8 = 6
1) 4 8/15 - 1 1/3 = 4 8/15 - 1 5/15 = 3 3/15 = 3 1/5
2) 3 1/5 * 1 7/8 = 16/5 * 15/8 = 6
(2 2/3 + 1 5/6) : 1 1/2 = 3
1) 2 2/3 + 1 5/6 = 2 4/6 + 1 5/6 = 3 9/6 = 4 3/6 = 4 1/2
2) 4 1/2 : 1 1/2 = 9/2 : 3/2 = 9/2 * 2/3 = 3
(3 1/6 - 2 7/15) : 1 2/5 = 1/2
1) 3 1/6 - 2 7/15 = 3 5/30 - 2 14/30 = 2 35/30 - 2 14/30 = 21/30 = 7/10
2) 7/10 : 1 2/5 = 7/10 : 7/5 = 7/10 * 5/7 = 1/2
(1 2/3 + 2 4/9) : (4 26/27 - 2 2/9) = 1 1/2
1) 1 2/3 + 2 4/9 = 1 6/9 + 2 4/9 = 3 10/9 = 4 1/9
2) 4 26/27 - 2 2/9 = 4 26/27 - 2 6/27 = 2 20/27
3) 4 1/9 : 2 20/27 = 37/9 : 74/27 = 37/9 * 27/74 = 3/2 = 1 1/2
(6 1/24 - 2/3) : (3 1/2 + 1 7/8) = 1
1) 6 1/24 - 2/3 = 6 1/24 - 16/24 = 5 25/24 - 16/24 = 5 9/24 = 5 3/8
2) 3 1/2 + 1 7/8 = 3 4/8 + 1 7/8 = 4 11/8 = 5 3/8
3) 5 3/8 : 5 3/8 = 43/8 : 43/8 = 43/8 * 8/43= 1
b1, q ---геометрическая прогрессия
b2 = b1*q, b3 = b1*q^2, b4 = b1*q^3
а1, d ---арифметическая прогрессия
a2 = a1+d, a3 = a1+2d, a4 = a1+3d
a1=b1, a2=b2+4, a3=b3+5, a4=b4
a2=b2+4 => b1+d = b1*q+4
a3=b3+5 => b1+2d = b1*q^2+5
a4=b4 => b1+3d = b1*q^3
система из трех уравнений...
из первого: b1*(q-1) = d-4
из третьего вычтем второе: d = b1*q^2*(q-1)-5 = q^2*(d-4)-5
q^2*(d-4) = d+5 => q^2 = (d+5) / (d-4)
из второго: b1-b1*q^2 = 5-2d
b1*(1-q^2) = 5-2d
b1*(1- (d+5) / (d-4) ) = 5-2d
b1*( 9 / (4-d) ) = 5-2d
b1 = (5-2d)(4-d) / 9
подставим в первое: (5-2d)(4-d) * (q-1) = 9(d-4)
q-1 = 9 / (2d-5)
q = 9 / (2d-5) + 1 = (2d+4) / (2d-5) => q^2 = (2d+4)^2 / (2d-5)^2 =>
(2d+4)^2 / (2d-5)^2 = (d+5) / (d-4)
(2d+4)^2 * (d-4) = (2d-5)^2 * (d+5) отсюда d = 7
q = (2*7+4) / (2*7-5) = 18/9
q = 2
a1 = b1 = (5-2*7)(4-7) / 9 = 3
геометрическая прогрессия: 3, 6, 12, 24
арифметическая прогрессия: 3, 10, 17, 24
6+4 = 10... 12+5 = 17...