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

To find the value of \( p \), we start by recognizing the property of a geometric series where the ratio \( r \) between consecutive terms is constant. Thus, we can express this relationship as follows: \[ \frac{u_2}{u_1} = \frac{u_3}{u_2} \] Substituting the values of \( u_1, u_2, \) and \( u_3 \): \[ \frac{q(2p + 3)}{q(4p + 1)} = \frac{q(2p - 3)}{q(2p + 3)} \] Removing \( q \) gives us: \[ \frac{2p + 3}{4p + 1} = \frac{2p - 3}{2p + 3} \] We cross-multiply to eliminate the fractions: \[ (2p + 3)^2 = (4p + 1)(2p - 3) \] Expanding both sides: \[ 4p^2 + 12p + 9 = 8p^2 - 12p + 2p - 3 \] Combining and simplifying terms results in: \[ 4p^2 - 26p + 12 = 0 \] Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ p = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 4 \cdot 12}}{2 \cdot 4} \] Calculating that we find: \[ p = \frac{26 \pm \sqrt{676 - 192}}{8} = \frac{26 \pm \sqrt{484}}{8} = \frac{26 \pm 22}{8} \] Calculating both roots gives: 1. \( p = \frac{48}{8} = 6 \) 2. \( p = \frac{4}{8} = \frac{1}{2} \) Since we need \( p \) as an integer, we take \( p = 6 \). Now, substituting back to find \( q \): The sum to infinity of a geometric series is given by \( S = \frac{u_1}{1 - r} \), and we know \( S = 250 \). First, we find \( r \): \[ r = \frac{u_2}{u_1} = \frac{q(2(6) + 3)}{q(4(6) + 1)} = \frac{q(12 + 3)}{q(24 + 1)} = \frac{15}{25} = \frac{3}{5} \] Substituting \( u_1 \) into the formula for \( S \): \[ 250 = \frac{q(4(6) + 1)}{1 - \frac{3}{5}} = \frac{q(24 + 1)}{\frac{2}{5}} = \frac{25q}{\frac{2}{5}} = \frac{125q}{2} \] Solving for \( q \): \[ 250 = \frac{125q}{2} \] \[ 500 = 125q \implies q = 4 \] Thus, the final values are: a) \( p = 6 \) and b) \( q = 4 \).