1. The following sequence of natural numbers is given, written in descending order:

....., $26,23,20, \ldots .$.

a) Find the first two terms of the sequence written in ascending order;

b) Find the sum of the first 20 terms of the sequence;

c) Prove that no term of the sequence can be a perfect square.
1.a) We observe that $26=3 \cdot 9-1 ; 23=3 \cdot 8-1 ; 20=3 \cdot 7-1$ and so on, we have that $2=3 \cdot 1-1$, so the first two terms of the sequence are 2 and $5(=3 \cdot 2-1)$

2 points.

b) The 20th term is $3 \cdot 20-1=59$

$S=2+5+8+\ldots+59=(2+59) \times 20: 2=610$

1 point.

c) Any term of the sequence is of the form $3n-1$. If it were a perfect square, then $3n-1=p^{2}$, so $3n=p^{2}+1$. If $p=3k$ then $3n=(3k)^{2}+1$, so $M_{3}=M_{3}+1$ which is false, if $p=3k+1$ then $3n=(3k+1)^{2}+1$, so $M_{3}=M_{3}+2$ which is false, if $p=3k+2$ then $3n=(3k+2)^{2}+1$, so $M_{3}=M_{3}+5$ which is false, in conclusion no term of the sequence can be a perfect square.

4 points.