Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $Ax^2 + Bx + C = 0$ with roots $\alpha, \beta$:
  • Sum of roots: $\alpha + \beta = -B/A$
  • Product of roots: $\alpha \beta = C/A$
• Reciprocal Roots: If $\alpha, \beta$ are roots of $Ax^2 + Bx + C = 0$, then $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $Cx^2 + Bx + A = 0$. The sum of reciprocal roots is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-B/A}{C/A} = -B/C$. The product of reciprocal roots is $\frac{1}{\alpha}\frac{1}{\beta} = \frac{1}{\alpha\beta} = A/C$.
• Arithmetic Progression (A.P.): A sequence $a_1, a_2, a_3, a_4$ is in A.P. if $a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d$ (common difference). This implies $a_1+a_4 = a_2+a_3$.

Step-by-Step Solution

Step 1: Analyze the first quadratic equation and its reciprocal roots. The first equation is $x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$. Let its roots be $p$ and $r$. Using Vieta's formulas: Sum of roots: $p+r = -(-8a)/1 = 8a$. Product of roots: $pr = 2a/1 = 2a$. The reciprocals of these roots are $\frac{1}{p}$ and $\frac{1}{r}$. Sum of reciprocal roots: $\frac{1}{p} + \frac{1}{r} = \frac{p+r}{pr} = \frac{8a}{2a} = 4$. (Equation 1) Product of reciprocal roots: $\frac{1}{p} \cdot \frac{1}{r} = \frac{1}{pr} = \frac{1}{2a}$. (Equation 2)

Step 2: Analyze the second quadratic equation and its reciprocal roots. The second equation is $x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$. Let its roots be $q$ and $s$. Using Vieta's formulas: Sum of roots: $q+s = -(12b)/1 = -12b$. Product of roots: $qs = 6b/1 = 6b$. The reciprocals of these roots are $\frac{1}{q}$ and $\frac{1}{s}$. Sum of reciprocal roots: $\frac{1}{q} + \frac{1}{s} = \frac{q+s}{qs} = \frac{-12b}{6b} = -2$. (Equation 3) Product of reciprocal roots: $\frac{1}{q} \cdot \frac{1}{s} = \frac{1}{qs} = \frac{1}{6b}$. (Equation 4)

Step 3: Use the A.P. condition to relate the reciprocal roots. We are given that $\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}$ are in A.P. Let these terms be $A_1, A_2, A_3, A_4$ respectively. So, $A_1 = \frac{1}{p}$, $A_2 = \frac{1}{q}$, $A_3 = \frac{1}{r}$, $A_4 = \frac{1}{s}$. From Equation 1, we have $A_1 + A_3 = 4$. From Equation 3, we have $A_2 + A_4 = -2$. Since $A_1, A_2, A_3, A_4$ are in A.P., we know that $A_1 + A_4 = A_2 + A_3$. Also, for an A.P., the sum of the first and last term equals the sum of the second and third term: $A_1 + A_4 = A_2 + A_3$. We also know that $A_1 + A_3 = 4$ and $A_2 + A_4 = -2$. Let the common difference be $d$. Then $A_2 = A_1+d$, $A_3 = A_1+2d$, $A_4 = A_1+3d$. Substituting these into the sum equations: $A_1 + (A_1+2d) = 4 \implies 2A_1+2d = 4 \implies A_1+d = 2$. (Equation 5) $(A_1+d) + (A_1+3d) = -2 \implies 2A_1+4d = -2 \implies A_1+2d = -1$. (Equation 6)

Step 4: Solve the system of equations for $A_1$ and $d$. Subtract Equation 5 from Equation 6: $(A_1+2d) - (A_1+d) = -1 - 2$ $d = -3$. Substitute $d=-3$ into Equation 5: $A_1 + (-3) = 2$ $A_1 = 5$. So, the terms of the A.P. are: $A_1 = \frac{1}{p} = 5$ $A_2 = \frac{1}{q} = A_1+d = 5-3 = 2$ $A_3 = \frac{1}{r} = A_1+2d = 5+2(-3) = 5-6 = -1$ $A_4 = \frac{1}{s} = A_1+3d = 5+3(-3) = 5-9 = -4$

Step 5: Calculate $a^{-1}$. From Equation 2, we have $\frac{1}{p} \cdot \frac{1}{r} = \frac{1}{2a}$. Using the values from Step 4: $A_1 \cdot A_3 = (5) \cdot (-1) = -5$. So, $\frac{1}{2a} = -5$. This implies $2a = -\frac{1}{5}$, so $a = -\frac{1}{10}$. Therefore, $a^{-1} = -10$.

Step 6: Calculate $b^{-1}$. From Equation 4, we have $\frac{1}{q} \cdot \frac{1}{s} = \frac{1}{6b}$. Using the values from Step 4: $A_2 \cdot A_4 = (2) \cdot (-4) = -8$. So, $\frac{1}{6b} = -8$. This implies $6b = -\frac{1}{8}$, so $b = -\frac{1}{48}$. Therefore, $b^{-1} = -48$.

Step 7: Calculate $a^{-1} - b^{-1}$. $a^{-1} - b^{-1} = (-10) - (-48) = -10 + 48 = 38$.

Common Mistakes & Tips

• Ensure the order of terms in the A.P. is correctly assigned to the reciprocal roots as given in the problem statement.
• Double-check the algebraic manipulations when solving the system of equations for the A.P. terms.
• Remember that $a$ and $b$ are non-zero, which implies $2a \neq 0$ and $6b \neq 0$, so the products of roots are non-zero, and their reciprocals are well-defined.

Summary We utilized Vieta's formulas to establish relationships between the reciprocal roots and the coefficients $a$ and $b$. The condition that the reciprocal roots form an arithmetic progression allowed us to set up and solve a system of linear equations to find the values of these reciprocal roots. Subsequently, using the product of reciprocal roots, we determined the values of $a^{-1}$ and $b^{-1}$, and finally computed their difference.

The final answer is $\boxed{38}$.