I found an answer from codegolf.stackexchange.com

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This is essentially the dead-end rule from snippet **3**: if there is only **one** ... Any
number from 0 to **9** would do pretty much the same. ... degrees, as opposed to
more **mathematical** radians of many languages use. ... Recall (from snippet **3**)
that QBasic outputs **text** to a window, not an output stream. ... 65 66 67 49 50 51
**27** 13 **9** 8.

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I found an answer from en.wikipedia.org

**Geometric progression** - Wikipedia

In **mathematics**, a **geometric progression**, also known as a **geometric sequence**,
is a sequence of numbers where each **term** after the first is found by multiplying ...

For more information, see **Geometric progression** - Wikipedia

Given geometric sequence

\frac{1}{3},\ \frac{1}{9},\ \frac{1}{27},......................

nth term = \frac{1}{19683}

First term a=\frac{1}{3}

Common ratio r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{3}{9}\ =\ \frac{1}{3}

Number of terms = n

nth term in geometric sequence a_n = ar^{n - 1}

a_n=ar^{n-1}=\frac{1}{19683}

\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}=\frac{1}{19683}

(\frac{1}{3})^{n-1\ +\ 1}=\frac{1}{19683}

\left(\frac{1}{3}\right)^n=\frac{1}{19683}

\left(\frac{1}{3}\right)^n=\left(\frac{1}{3^9}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \

Equating the powers

n = 9

Hence, \frac{1}{19683} is 9th term of Geometric sequence