Three numbet are in the ratio of 1:4:13. If 1 is added to all three number then they form geometric progression. Find the original numbers.

Answer 1

let the original numbers be x , 4x and 13x

(4x+1)/(x+1) = (13x+1)/(4x+1)
(4x+1)^2 = (x+1)(13x+1)
16x^2 + 8x + 1 = 13x^2 + 14x + 1
3x^2 - 6x = 0
3x(x - 2) = 0
x = 0 or x = 2
we'll reject the x = 0
so the 3 numbers are 2, 8 and 26

check: after adding 1 to each, the numbers would be 3, 9, and 27
which would be in a GP

If we accept 0,0,0, then add 1 to each, we get
1,1,1, which technically satisfy the definition of a GP, with r = 1
but in most definition for a GP, r ≠ 1
besides that, are 0,0,0 in the ratio of 1:4:13 ??

My conclusion is that the original numbers are 2, 8, and 26

Answer 2

let n = 1st number

(4n + 1) / (n + 1) = (13n + 1) / (4n + 1)

(4n + 1)^2 = (13n + 1)(n + 1) ... 16 n^2 + 8 n + 1 = 13 n^2 + 14 n + 1

3 n^2 - 6 n = 0 ... n^2 - 2n = 0 ... n = 0 , n = 2