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Arithmetic Progression

In the contin...

Question

In the continued fraction $\frac{b_{1}}{a_{1} -} \frac{b_{2}}{a_{2} -} \frac{b_{3}}{a_{3} -} \dots$ , if the denominator of every component exceed the numerator by unity at least, show that $p_{n}$ and $q_{n}$ increase with $n$ .

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Solution

$\Rightarrow \frac{b_{1}}{a_{1} -} \frac{b_{2}}{a_{2} -} \frac{b_{3}}{a_{3} -} . . . .$
$n^{t h}$ convergent can be given as;-
$p_{n} = a_{n} p_{n - 1} - b_{n} p_{n - 2}$
$q_{n} = a_{n} q_{n - 1} - b_{n} q_{n - 2}$
$∴ p_{n} - p_{n - 1} = (a_{n} - 1) p_{n - 1} - b_{n} p_{n - 2}$
Now $a_{n} - 1$ is atleast as great as $b_{n}$ ; therefore $p_{n} - p_{n - 1}$ is atleast as great as $b_{n} (p_{n - 1} - p_{n - 2})$ ; therefore $p_{n} > p_{n - 1}$ if $p_{n - 1} > p_{n - 2}$ ; so on . But $p_{2}$ is clearly greater than $p_{1}$ ; hence $p_{n} > p_{n - 1}$
Similarly $q_{n} > q_{n - 1}$

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\frac{b_{1}}{a_{1}} - \frac{b_{2}}{a_{2}} - \frac{b_{3}}{a_{3}} - \dots

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2 y - x = 8; 5 y - x = 14, y - 2 x = 1

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∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

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