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Find Four Numbers that form a geometric progression such that the second term is less than the first by 35, and the third term is greater than the fourth term by 560. (Note: there should be 3 different possible geometric sequences)

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Mhanifaidn

Answer:

  • -35/3, -140/3, - 560/3, -2240/3
  • 7, -28, 112, - 448  

Step-by-step explanation:

The terms are:

  • a₁, a₂, a₃, a₄

Given:

  • a₁ = a₂ + 35
  • a₃ = a₄ + 560

Use the nth term formula:

  • a₂ = a₁r
  • a₃ = a₁r²
  • a₄ = a₁r³

Substitute:

  • a₁ = a₁r + 35 ⇒ a₁(1 - r) = 35
  • a₁r² = a₁r³ + 560  ⇒ a₁(1 - r)r² = 560

Divide the second equation by the first:

  • r² = 560/35
  • r² = 16
  • r = √16
  • r = ± 4

Use the first equation to find the first term:

  • a₁( 1 ± 4) = 35
  • 1. a₁ = 35/-3 = -35/3
  • 2. a₁ = 35/5 = 7

We have two sequences:

r = 4

  • -35/3, -140/3, - 560/3, -2240/3

r = -4

  • 7, -28, 112, - 448  
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