Now, the player plays 10 more matches. Let’s analyze two scenarios:
So, scoring 1 additional goal increases the average from 0.15 to 0.20, which means an increase of:
\[ 0.20 - 0.15 = 0.05 \]
This increase of 0.05 in average due to 1 extra goal implies:
\[ \frac{1 \text{ goal}}{x + 10 \text{ matches}} = 0.05 \Rightarrow x + 10 = \frac{1}{0.05} = 20 \]
Hence, total matches after 10 more games = 20.
\[ \Rightarrow x = 20 - 10 = 10 \]
Number of matches already played = 10
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: