Solution
The first three terms of the geometric series $G$ are:
$$ U_1 = q(4p + 1), \quad U_2 = q(2p + 3), \quad U_3 = q(2p - 3). $$
(a) Find the possible values of $p$
The common ratio $r$ is:
$$ r = \frac{U_2}{U_1} = \frac{2p + 3}{4p + 1}, \quad r = \frac{U_3}{U_2} = \frac{2p - 3}{2p + 3}. $$
Equating the two expressions for $r$:
$$ \frac{2p + 3}{4p + 1} = \frac{2p - 3}{2p + 3}. $$
Cross-multiply:
$$ (2p + 3)^2 = (4p + 1)(2p - 3). $$
Step 1: Expand both sides
Expand $(2p + 3)^2$:
$$ (2p + 3)^2 = 4p^2 + 12p + 9. $$
Expand $(4p + 1)(2p - 3)$:
$$ (4p + 1)(2p - 3) = 8p^2 - 12p + 2p - 3 = 8p^2 - 10p - 3. $$
Equating the two sides:
$$ 4p^2 + 12p + 9 = 8p^2 - 10p - 3. $$
Step 2: Simplify the equation
Bring all terms to one side:
$$ 4p^2 + 12p + 9 - 8p^2 + 10p + 3 = 0. $$
$$ -4p^2 + 22p + 12 = 0. $$
Divide through by $-2$ for simplicity:
$$ 2p^2 - 11p - 6 = 0. $$
Step 3: Solve the quadratic equation
The quadratic equation is:
$$ 2p^2 - 11p - 6 = 0. $$
Use the quadratic formula:
$$ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$
Here, $a = 2$, $b = -11$, and $c = -6$. Substitute:
$$ p = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(2)(-6)}}{2(2)}. $$
$$ p = \frac{11 \pm \sqrt{121 + 48}}{4}. $$
$$ p = \frac{11 \pm \sqrt{169}}{4}. $$
$$ p = \frac{11 \pm 13}{4}. $$
Solve for the two possible values of $p$:
$$ p = \frac{11 + 13}{4} = \frac{24}{4} = 6, \quad p = \frac{11 - 13}{4} = \frac{-2}{4} = -\frac{1}{2}. $$
Final Answer:
$$ p = 6 \quad \text{or} \quad p = -\frac{1}{2}. $$
(b) Find the value of $q$
We are given that the sum to infinity $S_\infty$ of the convergent geometric series is 250. For a geometric series, the sum to infinity is given by:
$$ S_\infty = \frac{U_1}{1 - r}. $$
From part (a), we have $U_1 = q(4p + 1)$ and the common ratio
$$ r = \frac{2p + 3}{4p + 1}. $$
Case 1: $p = -\frac{1}{2}$
Substitute $p = -\frac{1}{2}$ into the expression for $r$:
$$ r = \frac{2p + 3}{4p + 1} = \frac{-1 + 3}{-2 + 1} = 2 $$
This contradicts convergence since $|r| > 1$.
Case 2: $p = 6$
Substitute $p = 6$ into the expression for $r$:
$$ r = \frac{2p + 3}{4p + 1} = \frac{15}{25} = 0.6 $$
Substitute these into the formula for $S_\infty$:
$$ 250 = \frac{q(4p + 1)} {1 - \frac{2p + 3}{4p + 1}} = \frac{25q}{1 - 0.6} $$
$$ 250 = \frac{25q}{0.4} $$
$$ q = 4 $$
Final Answer:
$$ q = 4 $$
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