‘A geometric progression has first term a and common ratior, and the terms are all different. The first, second and fourth terms of the geometricprogression form the first three terms of an arithmetic progression’.
Show that r³ - 2r + 1 =0
For the geometric bit;
U₁ = a
U₂ = ar
U₄ = ar³
For the arithmetic bit;
U₁ = a
U₂ = a + d
U₃ = a + 2d
So we can say:
ar³ = a + 2d and ar = a + d
d = ar – a
so 2d = 2ar – 2a
ar³ = 2ar – 2a
ar³ - 2ar + 2a = 0
r³ - 2r + 2 = 0 How is it +1, where does this come from??
Show that r³ - 2r + 1 =0
For the geometric bit;
U₁ = a
U₂ = ar
U₄ = ar³
For the arithmetic bit;
U₁ = a
U₂ = a + d
U₃ = a + 2d
So we can say:
ar³ = a + 2d and ar = a + d
d = ar – a
so 2d = 2ar – 2a
ar³ = 2ar – 2a
ar³ - 2ar + 2a = 0
r³ - 2r + 2 = 0 How is it +1, where does this come from??