Question:

Step 1: Set up the roots in arithmetic progression. Since the four roots are in arithmetic progression, let them be represented as $a-3d$, $a-d$, $a+d$, and $a+3d$, where $a$ is the center and $d$ is the common difference. Step 2: Use Vieta's formula for the sum of roots. By Vieta's formulas, the sum of all roots equals the negative of the coefficient of $x^3$ divided by the leading coefficient. Since there is no $x^3$ term in the polynomial $x^4 + qx^2 + kx + 225 = 0$, the coefficient of $x^3$ is $0$. $$\text{Sum of roots} = (a-3d) + (a-d) + (a+d) + (a+3d) = 4a = 0$$ Therefore, $a = 0$. Step 3: Identify the roots. With $a = 0$, the four roots simplify to: $$-3d, \quad -d, \quad d, \quad 3d$$ Step 4: Use Vieta's formula for the product of roots. By Vieta's formulas, the product of all roots equals the constant term (since the leading coefficient is $1$). $$(-3d)(-d)(d)(3d) = 9d^4 = 225$$ Solving for $d^2$: $$d^4 = 25 \implies d^2 = 5$$ Step 5: Calculate the sum of products of roots taken two at a time. By Vieta's formulas, the sum of products of roots taken two at a time equals the coefficient of $x^2$, which is $q$. $$q = (-3d)(-d) + (-3d)(d) + (-3d)(3d) + (-d)(d) + (-d)(3d) + (d)(3d)$$ Evaluating each term: $$q = 3d^2 - 3d^2 - 9d^2 - d^2 - 3d^2 + 3d^2$$ Step 6: Simplify and substitute the value of $d^2$. Combining like terms: $$q = (3 - 3 - 9 - 1 - 3 + 3)d^2 = -10d^2$$ Substituting $d^2 = 5$: $$q = -10(5) = -50$$ **Final Answer:** The value of $q$ is $\boxed{-50}$, which corresponds to **Option 4**.