Question

Aflati ratia si primul termen al progresiei geometrice (bn)n>= 1 stiind ca b1-b5=15 si b1-b3=7​

Answer 1

Răspuns:

b₁ = -49 ; q₁,₂ = ± 2√14 /7

Explicație pas cu pas:

{b1-b5=15

{b1-b3=7​  <=>

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{b₁-b₁q⁴ = 15

{b₁-b₁q² = 7 <=>

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{b₁(1-q⁴) = 15  

{b₁(1-q²) = 7   <=>

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{{b₁(1-q²)(1+q²) = 15    (1)

{b₁(1-q²) = 7  =>          (2)

din (1):(2) => 1+q² = 15/7 =>

q² = (15-7)/7 = 8/7 = 56/49 =>

q₁,₂ = ± 2√14 /7

q₁ = -2√14 /7

b₁ = 7/(1-q²) = 7/(1-8/7) =7/ (-1/7) = -49

b₁ = -49

b₂ = -49 · (-2√14/7) = 14√14

b₃ = 14√14  · (-2√14/7) = - 28·14/7 = -56

b₄ = -56  · (-2√14/7) = 112√14 /7 = 16√14

b₅ = 16√14 · (-2√14/7) = -32·14/7 = -64

b₁-b₅ = -49+64 = 15

b₁-b₃ = -49+56 = 7 , corect

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q₂ = 2√14 /7  ; b₁ = -49 =>

b₂ = -49 · 2√14 /7 = -14√14

b₃ = -14√14·2√14 /7 = -56

b₄ = -56· 2√14 /7 = -16√14

b₅ = -16√14 · 2√14 /7 = -64

b₁-b₅ = -49+64 = 15

b₁-b₃ = -49+56 = 7 , corect