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Step-by-Step Solution
Step 1: Write down the given quadratic equations and their roots
We have two quadratic equations:
• x^2 - 8ax + 2a = 0 with roots p and r .
• x^2 + 12bx + 6b = 0 with roots q and s .
Step 2: Note the condition on the reciprocals
It is given that
\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}
are in an Arithmetic Progression (A.P.). Four numbers are in A.P. if the difference between consecutive terms is constant.
Step 3: Form the equations whose roots are the reciprocals
To find the equation whose roots are the reciprocals of p and r , replace x with \frac{1}{x} in
x^2 - 8ax + 2a = 0,
then multiply through by x^2 . This gives:
1 - \frac{8a}{x} + \frac{2a}{x^2} = 0.
Multiplying both sides by x^2 :
2a\,x^2 - 8a\,x + 1 = 0.
So, the roots of 2a\,x^2 - 8a\,x + 1 = 0 are \frac{1}{p} and \frac{1}{r} .
Similarly, for x^2 + 12bx + 6b = 0 with roots q and s , replace x with \frac{1}{x} and multiply by x^2 :
1 + \frac{12b}{x} + \frac{6b}{x^2} = 0.
Multiplying both sides by x^2 :
6b\,x^2 + 12b\,x + 1 = 0.
Thus, the roots of 6b\,x^2 + 12b\,x + 1 = 0 are \frac{1}{q} and \frac{1}{s} .
Step 4: Express the A.P. condition explicitly
Four numbers in A.P. can be written as
\alpha - 3\beta,\; \alpha - \beta,\; \alpha + \beta,\; \alpha + 3\beta,
where the common difference is 2\beta . Hence,
\frac{1}{p} = \alpha - 3\beta,\quad
\frac{1}{q} = \alpha - \beta,\quad
\frac{1}{r} = \alpha + \beta,\quad
\frac{1}{s} = \alpha + 3\beta.
Step 5: Use the sum and product of reciprocals from the transformed quadratics
From 2a\,x^2 - 8a\,x + 1 = 0 whose roots are \frac{1}{p} and \frac{1}{r} :
• Sum of roots:
\frac{1}{p} + \frac{1}{r} = \frac{8a}{2a} = 4.
• Product of roots:
\left(\frac{1}{p}\right)\left(\frac{1}{r}\right) = \frac{1}{2a}.
From 6b\,x^2 + 12b\,x + 1 = 0 whose roots are \frac{1}{q} and \frac{1}{s} :
• Sum of roots:
\frac{1}{q} + \frac{1}{s} = -\frac{12b}{6b} = -2.
• Product of roots:
\left(\frac{1}{q}\right)\left(\frac{1}{s}\right) = \frac{1}{6b}.
Step 6: Relate the sums of reciprocals to the A.P. form
From the A.P. representation:
\frac{1}{p} + \frac{1}{r} = (\alpha - 3\beta) + (\alpha + \beta) = 2\alpha - 2\beta,
\frac{1}{q} + \frac{1}{s} = (\alpha - \beta) + (\alpha + 3\beta) = 2\alpha + 2\beta.
We know:
\frac{1}{p} + \frac{1}{r} = 4,\quad \frac{1}{q} + \frac{1}{s} = -2.
Therefore,
2\alpha - 2\beta = 4 \quad \Longrightarrow \quad \alpha - \beta = 2,
2\alpha + 2\beta = -2 \quad \Longrightarrow \quad \alpha + \beta = -1.
Adding these two equations:
(\alpha - \beta) + (\alpha + \beta) = 2 + (-1) \quad \Longrightarrow \quad 2\alpha = 1 \quad \Longrightarrow \quad \alpha = \frac{1}{2}.
From \alpha + \beta = -1 , we get \beta = -\frac{3}{2} .
Step 7: Determine each reciprocal
Substitute \alpha = \frac{1}{2} , \beta = -\frac{3}{2} back into the A.P. forms:
\frac{1}{p} = \alpha - 3\beta
= \frac{1}{2} - 3\left(-\frac{3}{2}\right)
= \frac{1}{2} + \frac{9}{2}
= 5,
\frac{1}{q} = \alpha - \beta
= \frac{1}{2} - \left(-\frac{3}{2}\right)
= \frac{1}{2} + \frac{3}{2}
= 2,
\frac{1}{r} = \alpha + \beta
= \frac{1}{2} + \left(-\frac{3}{2}\right)
= -1,
\frac{1}{s} = \alpha + 3\beta
= \frac{1}{2} + 3\left(-\frac{3}{2}\right)
= \frac{1}{2} - \frac{9}{2}
= -4.
Step 8: Use products of the reciprocals to find a and b
From \frac{1}{p} \cdot \frac{1}{r} = \frac{1}{2a} = 5 \times (-1) = -5 , we get
\frac{1}{2a} = -5 \quad \Longrightarrow \quad \frac{1}{a} = -10.
From \frac{1}{q} \cdot \frac{1}{s} = \frac{1}{6b} = 2 \times (-4) = -8 , we get
\frac{1}{6b} = -8 \quad \Longrightarrow \quad \frac{1}{b} = -48.
Step 9: Compute a^{-1} - b^{-1}
We have \frac{1}{a} = -10 and \frac{1}{b} = -48 . Thus,
a^{-1} - b^{-1} = \frac{1}{a} - \frac{1}{b} = -10 - \bigl(-48\bigr) = 38.
Final Answer
a^{-1} - b^{-1} = 38.
