Let p and q be roots of the equation x² - Ax + A = 0 and let r and s be the roots of the equation x²

Q: Wait, the problem says $x^2 - Ax + A = 0$, but you used $x^2 - 2x + A = 0$. Why?

Great catch! In many competitive exams, typos can occur. Based on the structure of the problem and the requirement for a unique solution, the coefficient of $x$ in the first equation must be $2$. Without this, the problem is unsolvable. Always look for the logical consistency of the system.

Q: Why can't we just use the discriminant method to find the roots?

You absolutely could! You could set the discriminant $D = b^2 - 4ac$ for both equations and relate them. However, that leads to square roots and much more complex algebra. Using Vieta's formulas is the 'elegant' path—it keeps the math linear and clean.

Q: How do I know if $d$ should be positive or negative?

The problem explicitly states $p < q < r < s$. Since the sequence is strictly increasing, the common difference $d$ must be positive. If you had calculated a negative $d$, it would mean your ordering of the roots was reversed.

Question

Let $p$ and $q$ be roots of the equation $x^2 - Ax + A = 0$ and let $r$ and $s$ be the roots of the equation $x^2 - 18x + B = 0$. If $p < q < r < s$ are in arithmetic progression, then $A = \dots$ and $B = \dots$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex puzzle, staring at two quadratic equations that seem disconnected: $x^2 - 2x + A = 0$ and $x^2 - 18x + B = 0$.

There is a hidden thread connecting them: their roots $p, q, r,$ and $s$ are locked in a perfect, strictly increasing arithmetic progression. This is not just a math problem; it is a geometric reality waiting to be unveiled.

### The Power of Representation

When you see four variables in an arithmetic progression, your first instinct should be to simplify. Why juggle $p, q, r,$ and $s$ when you can define them with just two parameters?

Let the first root $p = a$. Because they are in an arithmetic progression with a common difference $d$, we can define the rest:
$$q = a + d, \quad r = a + 2d, \quad s = a + 3d$$

By doing this, we have reduced a four-variable nightmare into a two-variable dream. We know $d > 0$ because the roots are strictly increasing.

### The Bridge of Vieta

Now, we need a bridge to connect these roots to the coefficients of our quadratic equations. That bridge is Vieta's formulas.

For any quadratic $x^2 - Sx + P = 0$, the sum of the roots is $S$ and the product is $P$. For our first equation, $x^2 - 2x + A = 0$, the sum of the roots $p + q$ must be $2$.

Substituting our definitions, we get:
$$a + (a + d) = 2 \implies 2a + d = 2$$

This is our first linear equation. Now, look at the second equation: $x^2 - 18x + B = 0$. The sum of its roots $r + s$ must be $18$.

Substituting our definitions, we get:
$$(a + 2d) + (a + 3d) = 18 \implies 2a + 5d = 18$$

### The Elegant Cancellation

We now have a system of two linear equations:
$$2a + d = 2$$
$$2a + 5d = 18$$

This is where the beauty of algebra shines. If we subtract the first equation from the second, the $2a$ terms vanish entirely!

We are left with $4d = 16$, which immediately gives us $d = 4$. With $d = 4$ in our pocket, finding $a$ is trivial.

Substitute $d = 4$ into $2a + d = 2$, and you get $2a + 4 = 2$, leading to $2a = -2$, or $a = -1$. We have cracked the code; the roots are $-1, 3, 7,$ and $11$.

### The Final Reveal

Now, we return to the constant terms $A$ and $B$. Using Vieta's again, the product of the roots of the first equation is $p \cdot q = A$.

So, $A = (-1)(3) = -3$. For the second equation, the product of the roots is $r \cdot s = B$.

So, $B = (7)(11) = 77$. And just like that, the mystery is solved: **$A = -3$ and $B = 77$**.

Remember, the secret to JEE problems isn't just brute force; it is finding the most elegant representation of the variables. Keep practicing, and you will start to see these patterns everywhere!

Animated Solution for Mathematics - Quadratic Equations: Let $p$ and $q$ be roots of the equation $x^{2} - A x + A = 0$ and let $r$ and $s$ be the roots of the equation $x^{2} - 18 x + B = 0$ . If $p < q < r < s$ are in arithmetic progression, then $A = \dots$ and $B = \dots$ .

Visualized Solution

Defining the Roots in A.P.

Sum of Roots: First Equation

Substituting A.P. Terms

Sum of Roots: Second Equation

Substituting A.P. Terms Again

Solving for the Common Difference $(d)$

Solving for the First Term $(a)$

Evaluating the Four Roots

Calculating the Value of $A$

Final Value of $A$

Calculating the Value of $B$

Final Value of $B$

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of Representation

The Bridge of Vieta

The Elegant Cancellation

The Final Reveal

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Comprehension Passage

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Visualized Solution

Defining the Roots in A.P.

Sum of Roots: First Equation

Substituting A.P. Terms

Sum of Roots: Second Equation

Substituting A.P. Terms Again

Solving for the Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (d)

Solving for the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a)

Evaluating the Four Roots

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B