1) Fie a,b,c termenii consecutivi ai unei progresii aritmetice →
atunci 2b = a+c
2(2x-2) = x-1+x+3
4x-4 = 2x+2
4x-2x = 2+4
2x = 6
x = 3
2) Fie a,b,c termenii consecutivi ai unei progresii geometrice →
atunci b = √(a×c)
x+1 = √(x-1)(2x+3) ridici totul la a 2-a
(x+1)² = (x-1)(2x+3)
x²+2x+1 = 2x²+3x-2x-3
x²-x-4 = 0
Δ = 1+16 = 17
x₁ = (1+√17)/2
x₂ = (1-√17)/2
3) b₂ = b1 × q
12 = 3 × q
q = 4
b₅ = b₂ × q³ = 12 × 4³ = 12 × 64 = 768
