Question

Sa se determine termenul de rang n al progresiei geometrice Ș

a) 6, 18, 54, 162, ......

b) $\sqrt 6, \sqrt 3, \sqrt 6/2, ......$

Answer 1

Răspuns:

Explicație pas cu pas:

a) 6, 18, 54, 162, ......

b₁ = 6 ; b₂ = 18 ; b₂ = b₁q => q = b₂/b₁ = 18/6 => q = 3

b₃ = b₂q = b₁q² = 18·3 = 54

bₙ = b₁qⁿ⁻¹ = 6·3ⁿ⁻¹ = 6·3ⁿ : 3 = 2·3ⁿ

bₙ = 2·3ⁿ

b) √6 ; √3 ; √6/2 ; ......

b₁ = √6 ; b₂ = √3 = b₁q =>

q = b₂/b₁ = √3/√6 = √18/6 = 3√2 /6 = √2/2

q = √2/2

b₃ = b₂q = √3·√2/2 = √6/2

bₙ = b₁qⁿ⁻¹ = √6·(√2/2)ⁿ⁻¹ = √6·(2√2/2)·(√2/2)ⁿ

bₙ = √12·(√2/2)ⁿ = 2√3·(√2/2)ⁿ

bₙ = 2√3·(√2/2)ⁿ

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