JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Consider two sets $A$ and $B$ , each containing three numbers in A.P. Let the sum and the product of the elements of $A$ be 36 and $p$ respectively and the sum and the product of the elements of $B$ be 36 and $q$ respectively. Let $d$ and $D$ be the common differences of AP's in $A$ and $B$ respectively such that $D = d + 3, d > 0$ . If $\frac{p + q}{p - q} = \frac{19}{5}$ , then $p - q$ is equal to

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Progressive Hints

Hint 1

The Spark: When dealing with three numbers in an Arithmetic Progression, avoid the standard

a, a + d, a + 2 d

notation. Instead, use the symmetric form

a - d, a, a + d

. This choice makes the sum of the terms incredibly simple because the common difference

d

cancels out immediately.

Hint 2

The Arsenal: You have two sets,

A

and

B

, with products

p

and

q

. Express these products using the symmetric terms:

p = (12 - d) (12) (12 + d) = 12 (144 - d^{2})

and

q = 12 (144 - D^{2})

. The ratio

\frac{p + q}{p - q} = \frac{19}{5}

is a classic algebraic structure that simplifies significantly if you treat the terms

144 - d^{2}

and

144 - D^{2}

as single units.

Hint 3

The Execution: Start by substituting the expressions for

p

and

q

into the ratio

\frac{p + q}{p - q} = \frac{19}{5}

. You will find that the factor of

12

cancels out entirely. Then, substitute

D = d + 3

into the equation. This will leave you with a quadratic equation in

d

. Remember to check the condition

d > 0

when you solve for the roots.

Visualized Solution

Visualizing the Sets in A.P.

Consider Set $A = {a - d, a, a + d}$
Consider Set $B = {b - D, b, b + D}$
Common differences are $d$ and $D$ respectively.

Finding the Middle Term of Set $A$

Sum of elements in $A = 36$
$(a - d) + a + (a + d) = 36$
$3 a = 36 ⟹ a = 12$

Finding the Middle Term of Set $B$

Sum of elements in $B = 36$
$(b - D) + b + (b + D) = 36$
$3 b = 36 ⟹ b = 12$

Defining Products $p$ and $q$

Product of $A$ : $p = (12 - d) (12) (12 + d) = 12 (144 - d^{2})$
Product of $B$ : $q = (12 - D) (12) (12 + D) = 12 (144 - D^{2})$

Setting up the Ratio Equation

Given: $\frac{p + q}{p - q} = \frac{19}{5}$
Substitute $p$ and $q$ : $\frac{12 ( 144 - d ^{2} ) + 12 ( 144 - D ^{2} )}{12 ( 144 - d ^{2} ) - 12 ( 144 - D ^{2} )} = \frac{19}{5}$
Simplify by canceling $12$ : $\frac{288 - d ^{2} - D ^{2}}{D ^{2} - d ^{2}} = \frac{19}{5}$

Substituting $D = d + 3$

Given relation: $D = d + 3$
Substitute into the ratio: $\frac{288 - d ^{2} - ( d + 3 ) ^{2}}{( d + 3 ) ^{2} - d ^{2}} = \frac{19}{5}$
Denominator: $(d + 3)^{2} - d^{2} = 6 d + 9$
Numerator: $288 - d^{2} - (d^{2} + 6 d + 9) = 279 - 2 d^{2} - 6 d$

Forming the Quadratic Equation

Cross-multiply: $5 (279 - 2 d^{2} - 6 d) = 19 (6 d + 9)$
Expand: $1395 - 10 d^{2} - 30 d = 114 d + 171$
Rearrange terms: $10 d^{2} + 144 d - 1224 = 0$
Divide by 2: $5 d^{2} + 72 d - 612 = 0$

Solving for $d$

Factorize: $5 d^{2} + 102 d - 30 d - 612 = 0$
$(5 d + 102) (d - 6) = 0$
Since $d > 0$ , we must choose $d = 6$
Consequently, $D = 6 + 3 = 9$

Calculating the Final Answer $p - q$

We need to find $p - q$
$p - q = 12 (144 - d^{2}) - 12 (144 - D^{2})$
Simplify: $p - q = 12 (D^{2} - d^{2})$
Substitute $D = 9, d = 6$ : $p - q = 12 (9^{2} - 6^{2})$
$p - q = 12 (81 - 36) = 12 (45) = 540$

Key Takeaways

Symmetry is Power: Choosing terms as $(a - d, a, a + d)$ drastically simplifies sum equations.
Constraint Checking: Always verify conditions like $d > 0$ before finalizing roots.
Algebraic Foresight: Delaying the multiplication of constants (like $12$ ) often leads to elegant cancellations.

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The Sigma Insight: Arithmetic Progression (A.P.)

Solution Diagram

Analyzing the Setup

Imagine you are standing before a complex puzzle. You have two sets of numbers,

A

and

B

, both dancing in an Arithmetic Progression.

When we assume the terms of set

A

are

a - d, a, a + d

, we are choosing a perspective that makes the math collapse into simplicity. The sum of these terms is

36

, and because of our symmetric choice, the

d

terms vanish into thin air.

This leaves us with:

3 a = 36 \Rightarrow a = 12

The same logic applies to set

B

, where the middle term

b

is also

12

. We have already stripped away the first layer of complexity.

The Product Trap

Now, we move to the products

p

and

q

. The product of set

A

p = (12 - d) (12) (12 + d)

. Using the difference of squares, this becomes:

p = 12 (144 - d^{2})

Similarly, for set

B

, we have:

q = 12 (144 - D^{2})

Many students rush to multiply these out, but the key here is to keep the expression factored. Because we are given the ratio

\frac{p + q}{p - q} = \frac{19}{5}

, keeping the

12

factored allows us to cancel it out entirely from the numerator and denominator.

The Ratio Simplification

Let us look at the ratio equation:

\frac{12 ( 144 - d ^{2} ) + 12 ( 144 - D ^{2} )}{12 ( 144 - d ^{2} ) - 12 ( 144 - D ^{2} )} = \frac{19}{5}

After canceling the

12

, we are left with:

\frac{288 - d ^{2} - D ^{2}}{D ^{2} - d ^{2}} = \frac{19}{5}

Now, we introduce the constraint

D = d + 3

. Substituting this into the denominator gives

(d + 3)^{2} - d^{2} = 6 d + 9

. The numerator becomes

288 - d^{2} - (d^{2} + 6 d + 9) = 279 - 2 d^{2} - 6 d

The Final Victory

Cross-multiplying gives us:

5 (279 - 2 d^{2} - 6 d) = 19 (6 d + 9)

Expanding this leads to:

1395 - 10 d^{2} - 30 d = 114 d + 171

Rearranging everything to one side, we get:

10 d^{2} + 144 d - 1224 = 0

Dividing by

2

yields

5 d^{2} + 72 d - 612 = 0

. Factoring this quadratic, we find:

(5 d + 102) (d - 6) = 0

Since the problem demands

d > 0

, we reject the negative root and embrace

d = 6

. This implies

D = 9

Finally, we calculate the difference:

p - q = 12 (D^{2} - d^{2}) = 12 (81 - 36) = 12 (45) = 540

The final answer is 540.

Common Traps & FAQs

Why do we use

a - d, a, a + d

instead of

a, a + d, a + 2 d

Great question! While both methods are mathematically valid, using

a - d, a, a + d

is a powerful strategy in competitive exams. When you sum these terms, the

- d

and

+ d

cancel out, leaving you with

3 a = 36

, which gives

a = 12

instantly. If you used

a, a + d, a + 2 d

, you would get

3 a + 3 d = 36

, or

a + d = 12

, which is slightly more cumbersome to work with later.

Why can't we just solve for

p

and

q

individually first?

You could, but it would be a trap! The problem gives you the ratio

\frac{p + q}{p - q}

, not the individual values. If you try to find

p

and

q

first, you will be stuck with two unknowns (

d

and

D

) and only one ratio equation. By working with the ratio directly, you allow the constants to cancel out, which is a much more elegant and efficient path to the solution.

Wait, why do we reject the negative root for

d

Always look closely at the problem statement! The problem explicitly states that

d > 0

. In quadratic equations, we often get two roots, but in physics and math problems, we must always check if the roots satisfy the physical or logical constraints provided in the question. Since

d

must be positive, we discard any negative result.

Select Answer:

Visualized Solution

Visualizing the Sets in A.P.

Finding the Middle Term of Set $A$

Finding the Middle Term of Set $B$

Defining Products $p$ and $q$

Setting up the Ratio Equation

Substituting $D = d + 3$

Forming the Quadratic Equation

Solving for $d$

Calculating the Final Answer $p - q$

Key Takeaways

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Product Trap

The Ratio Simplification

The Final Victory

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Select Answer:

Visualized Solution

Visualizing the Sets in A.P.

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

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

Defining Products .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Setting up the Ratio Equation

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=d+3

Forming the Quadratic Equation

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

Calculating the Final Answer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p−q

Key Takeaways

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Product Trap

The Ratio Simplification

The Final Victory

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