\( u_{1}=q(4 p+1), u_{2}=q(2 p+3) \) and \( u_{3}=q(2 p-3) \) a) Find the value of \( p \) as an integer 1 The sum to infinity of the series is 250 . b) Find the value of \( q . \square I \)

To find the value of \( p \), we know that the sum to infinity \( S \) for a geometric series is given by \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Here, we can identify \( a \) as \( u_1 = q(4p + 1) \) and \( u_2 = q(2p + 3) \) to find the common ratio \( r \). Given \( S = 250 \), we can set up our equations accordingly. After deriving and simplifying the expressions based on the conditions provided, you’ll determine the integer value of \( p \) that satisfies the equation. For the value of \( q \), after finding the integer \( p \), substitute it back into \( u_1 \) and compute \( q \) from the equation \( 250 = \frac{q(4p + 1)}{1 - r} \) you've previously established. This way, you'll get \( q \) depending on the identified \( p \). Play around with different values for \( p \) to hit that sweet spot of 250, and voila! You’ve got your \( q \).