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 \( q \).

We are given that the first three terms of a geometric series are   u₁ = q(4p + 1)   u₂ = q(2p + 3)   u₃ = q(2p – 3) and that the sum to infinity of the series is 250. Step 1. Find p. For a geometric series the ratio of successive terms is constant. Thus, u₂/u₁ = u₃/u₂. Using the given terms, we have   [q(2p + 3)]² = [q(4p + 1)][q(2p – 3)]. Since q ≠ 0 we can cancel q²:   (2p + 3)² = (4p + 1)(2p – 3). Expanding both sides:   LHS: (2p + 3)² = 4p² + 12p + 9.   RHS: (4p + 1)(2p – 3) = 8p² – 10p – 3. Setting them equal gives:   4p² + 12p + 9 = 8p² – 10p – 3. Rearrange by subtracting the left side from the right side:   0 = 8p² – 10p – 3 – 4p² – 12p – 9   0 = 4p² – 22p – 12. Divide the entire equation by 2 to simplify:   0 = 2p² – 11p – 6. This quadratic factors or can be solved using the quadratic formula:   p = [11 ± √(11² – 4·2·(–6))] / (2·2)     = [11 ± √(121 + 48)] / 4     = [11 ± √169] / 4     = [11 ± 13] / 4. Thus, the two possible solutions for p are:   p = (11 + 13)/4 = 24/4 = 6  and  p = (11 – 13)/4 = –2/4 = –½. Since the problem asks for p as an integer, we choose p = 6. Step 2. Find q. Substitute p = 6 into the first three terms to verify the common ratio and then use the sum to infinity formula. Calculate the first three terms:   u₁ = q(4·6 + 1) = q(24 + 1) = 25q,   u₂ = q(2·6 + 3) = q(12 + 3) = 15q,   u₃ = q(2·6 – 3) = q(12 – 3) = 9q. The common ratio is:   r = u₂/u₁ = (15q)/(25q) = 3/5    (and u₃/u₂ = 9q/15q = 3/5). The sum to infinity for a geometric series is given by:   S∞ = u₁ / (1 – r). We are told S∞ = 250, so:   (25q) / (1 – 3/5) = 250. Since 1 – 3/5 = 2/5, the equation becomes:   25q / (2/5) = 250   ⟹  25q × (5/2) = 250   ⟹  (125/2)q = 250. Solve for q by multiplying both sides by 2/125:   q = 250 × (2/125) = 500/125 = 4. Final Answers: a) p = 6 b) q = 4.