Where Does \(\frac{3}{7}\) Appear for the 5th Time in the Rational Series?

[MarkFL]

Here is this week's POTW:

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The positive rational numbers may be arranged in the form of a simple series as follows:

\(\displaystyle \frac{1}{1},\,\frac{2}{1},\,\frac{1}{2},\,\frac{3}{1},\,\frac{2}{2},\,\frac{1}{3},\,\frac{4}{1},\,\frac{3}{2},\,\frac{2}{3},\,\frac{1}{4},\,\cdots\)

In this series, every rational number is repeated indefinitely. In which term (for example the second appearance of 1 is the 5th term) of this series will the value \(\displaystyle \frac{3}{7}\) appear for the 5th time?

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[MarkFL]

Congratulations to the following members for their correct solutions:

• greg1313
• kaliprasad
• lfdahl
• fatimarose23

The solution given by fatimarose23 is as follows:

Spoiler

We denote the sequence as below
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{1}=1/1,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{2}=2/1, \ \ \ \ a_3=1/2,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_4=3/1, \ \ \ a_5=2/2, \ \ a_6=1/3,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_7=4/1,\ \ \ \ a_8=3/2, \ \ \ a_9=2/3, \ \ \ a_{10}=1/4,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ a_{11}=5/1, \ \ \ a_{12}=4/2, \ \ \ a_{13}=3/3, \ \ \ a_{14}=2/4,\ \ \ a_{15}=1/5,\ \ \ \ \ \ \ \ \ \ \ $$ $$ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{16}=6/1,\ \ \ a_{17}=5/2, \ \ \ a_{18}=4/3, \ \ \ a_{19}=3/4, \ \ \ a_{20}=2/5, \ \ \ a_{21}=1/6,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

We know the summation $ S $ of first $ n $ integer numbers, that is $ 1+2+3+⋯+n $ obtained by $ S=(n(n+1))/2 $, by considering this point,
If $ a_{i}=n_1/n_2 $ and we want to obtain $ i $ via $ n_{1} , n_{2} $. First set $ α=n_{2}+n_{1}-1 $ then define $ s=(α(α+1))/2 $ at last $ i=s-n_{1}+1 $.
Now for fraction $ 3/7 $ we want to know which $ i $ shows it for the fifth time.
We know $ 3/7=6/14=9/21=12/28=15/35 $ hence the fifth time which we have $ 3/7 $ is the sentence $ 15/35 $, that is $ n_{1}= 15 ,n_{2}= 35 $
So
$$ α=15+35-1=49 $$
And

$$ s=(49\times50)/2=1225 $$
Then
$$ i=1225-15+1=1211$$
The answer is $ a_{1211 }=15/35 $ that shows $ 3/7 $ for the fifth time.

The solution given by greg1313 is very similar to my own:

Spoiler

By inspection, a formula for the $n$th term with numerator $p$ and denominator $q$ is $\dfrac{(p+q-1)(p+q)-2(p-1)}{2}$.

The list of terms occurring in the series that have value $\dfrac37$ are $\dfrac37,\dfrac{6}{14},\dfrac{9}{21},\dfrac{12}{28},\dfrac{15}{35}$.

Evaluating the formula given above with $p=15$ and $q=35$, the $5$th occurrence of the value $\dfrac37$ is at the $1211$th term.