2. Consider the sequence of natural numbers $3,10,17,24,31, \ldots$.

a) Determine the 2014th term of the sequence.

b) Determine the numbers $x$ and $y$ knowing that they are consecutive terms of the sequence and $x<608<y$.

Ionuţ Mazalu, Brăila
## Solution:

a) $3=7 \cdot 0+3,10=7 \cdot 1+3,17=7 \cdot 2+3, \ldots$ $2 p$

The 2014th term is equal to $7 \cdot 2013+3=14094$ $2 p$

b) $608=7 \cdot 602+6$ 1p

$x=7 \cdot 86+6-3=605$ and $y=7 \cdot 87+3=612$ $2 p$