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)

-35/3, -140/3, - 560/3, -2240/37, -28, 112, - 448  

Step-by-step explanation:

The terms are:

a₁, a₂, a₃, a₄

Given:

a₁ = a₂ + 35a₃ = a₄ + 560

Use the nth term formula:

a₂ = a₁ra₃ = a₁r²a₄ = a₁r³

Substitute:

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

Divide the second equation by the first:

r² = 560/35r² = 16r = √16r = ± 4

Use the first equation to find the first term:

a₁( 1 ± 4) = 351. a₁ = 35/-3 = -35/32. a₁ = 35/5 = 7

We have two sequences:

r = 4

-35/3, -140/3, - 560/3, -2240/3

r = -4

7, -28, 112, - 448  

What the heck i didn’t know