Arithmetic Progression Problem

[anemone]

Find three irreducible fractions $\dfrac{a}{d}$, $\dfrac{b}{d}$ and $\dfrac{c}{d}$ that form an arithmetic progression, if $\dfrac{b}{a}=\dfrac{1+a}{1+d}$, $\dfrac{c}{b}=\dfrac{1+b}{1+d}$.

 

[mente oscura]

Find three irreducible fractions $\dfrac{a}{d}$, $\dfrac{b}{d}$ and $\dfrac{c}{d}$ that form an arithmetic progression, if $\dfrac{b}{a}=\dfrac{1+a}{1+d}$, $\dfrac{c}{b}=\dfrac{1+b}{1+d}$.

Hello.

Spoiler

$$If \ a<d \rightarrow{}a>b>c$$

$$a<d \rightarrow a+1<d+1$$

Same "c".

$$k=a-b, k=$$reason of the arithmetic progression.

$$k=a-\dfrac{a^2+a}{d+1}=\dfrac{ad-a^2}{d+1}$$

$$b-k=c \rightarrow{}b-\dfrac{ad-a^2}{d+1}=\dfrac{b^2+b}{d+1}$$

Resolving:

$$a^2-ad-b^2+bd=0$$

$$a=\dfrac{d \pm \sqrt{d^2+4b^2-4bd}}{2}$$

$$a=\dfrac{d \pm \sqrt{(d-2b)^2}}{2}$$

$$a=d-b \ or \ a=b$$

$$a=d-b \rightarrow{}d=a+b$$

$$\dfrac{b}{a}=\dfrac{1+a}{1+a+b} \rightarrow{}a^2+a=b^2+b+ab$$

$$a=\dfrac{(b-1) \pm \sqrt{(b-1)^2+4b^2+4b}}{2}$$

$$(b-1)^2+4b^2+4b=5b^2+2b+1=T^2$$

$$For \ b=2, T=5$$

$$a=3, b=2, c=1, d=5$$

Regards.

 

[anemone]

Thanks for participating, mente oscura! Your answer is correct, bravo!