Question

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)

Answer 1

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