Question

b3=12
b3-b2=6
Sa aflu b20, S30
progresii geometrice​

Answer 1

Răspuns:

Explicație pas cu pas:

b₃ = 12

b₃-b₂ = 6

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b₃ = b₁·q² = 12  => b₁ = 12/q²

b₂ = b₁·q

b₁·q²-b₁·q = 6  <=> b₁·q·(q-1) = 6  =>

q²-q = 6/b₁ = 6/(12/q²) = q²/2 =>

q²-2q = 0  <=> q(q-2) = 0  => q₁ = 0 ; q₂ = 2

b₁ = 12/4 = 3

b₂ = 3·2 = 6

b₃ = 6·2 = 12

b₂₀ = b₁·q¹⁹ => b₂₀ = 3·2¹⁹

S₃₀ = b₁+b₂+....+b₃₀ = b₁·(1+q+q²+....+q²⁹)

S₃₀ = 3·(1+2+2²+......+2²⁹) = 3·a =>

2·a = 2+2²+2³+....+2²⁹+2³⁰ = a-1+2³⁰ => a = 2³⁰-1

S₃₀ = 3·(2³⁰-1)