przyjmiemy że ciąg (an) jest ciągiem geometrycznym . znajdź : a) iloraz tego ciągu  i a10 jeśli a1=1 i a11=32 b) a4, jesli a10=24 i iloraz  wynosi q =√2 c) a10 jesli  a13= (-1) i a5= (-19)  

a) a1 = 1  oraz a11 = 32 Mamy a11 = a1*q¹⁰ q¹⁰ = a¹¹ : a¹ = 32 : 1 = 32 q = √2 --------- a₁₀ = a1*q⁹ = 1 *(√2)⁹ = 16√2 =========================== b) a₁₀ = 24 q = √2 Mamy a1*q⁹ = 24 a1*(√2)⁹ = 24 a1*16√2 = 24 a1 = 24 : 16√2 = (3√2)/4 a₄ = a1*q³ =[ (3√2)/4]* (√2)³ = 3 ================================ c) a₁₃ = -1  oraz  a₅ = -1/9 q⁸ = -1 :(-1/9) = 9 q⁴ = 3 q = ⁴√3 a₅ = a1*q⁴ --> a₁ = a₅ : q⁴ = (-1/9) :3 = -1/27 a₁₀ = a₁*q⁰ = (-1/27) *(⁴√3)⁹ = (-1/27)*9*⁴√3 =( -1/3)*⁴√3 ===================================================