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.

Reiny Reiny answered
4 years ago

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

R_Scott R_Scott answered
4 years ago

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

Explain Bot Explain Bot answered
7 months ago

To solve this problem, let's start by assuming the three original numbers are x, 4x, and 13x.

According to the problem, if we add 1 to each of these numbers, they will form a geometric progression.

So, the new numbers will be x+1, 4x+1, and 13x+1.

A geometric progression means that each term is obtained by multiplying the previous term by a constant ratio.

Therefore, we can write:

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

To solve this equation, we can cross-multiply:

(4x+1) * (4x+1) = (x+1) * (13x+1)

Expanding both sides:

16x^2 + 8x + 1 = 13x^2 + 14x + 1

Simplifying:

16x^2 + 8x = 13x^2 + 14x

Bringing all terms to one side:

16x^2 - 13x^2 - 8x + 14x = 0

3x^2 + 6x = 0

Factoring out x:

x(3x + 6) = 0

Setting each factor equal to zero:

x = 0 or 3x + 6 = 0

If x = 0, then the three numbers are 0, 0, and 0, which does not satisfy the given condition that they are in the ratio of 1:4:13.

If 3x + 6 = 0, then 3x = -6, and x = -2.

So the original numbers are x = -2, 4x = -8, and 13x = -26.

Therefore, the original numbers are -2, -8, and -26.

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