MHB Arithmetic Progression Problem

AI Thread Summary
The discussion centers on finding three irreducible fractions, $\dfrac{a}{d}$, $\dfrac{b}{d}$, and $\dfrac{c}{d}$, that form an arithmetic progression under specific conditions. The conditions are given by the equations $\dfrac{b}{a}=\dfrac{1+a}{1+d}$ and $\dfrac{c}{b}=\dfrac{1+b}{1+d}$. Participants engage in solving the problem, with one user confirming the correctness of another's answer. The focus remains on deriving the fractions that satisfy the arithmetic progression criteria. The thread emphasizes the mathematical relationships between the fractions involved.
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Find three irreducible fractions $\dfrac{a}{d}$, $\dfrac{b}{d}$ and $\dfrac{c}{d}$ that form an arithmetic progression, if $\dfrac{b}{a}=\dfrac{1+a}{1+d}$, $\dfrac{c}{b}=\dfrac{1+b}{1+d}$.
 
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anemone said:
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.

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.
 
Last edited:
Thanks for participating, mente oscura! Your answer is correct, bravo!
 
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