Maths Q6 – JEE-MAIN 2025

Home Maths Sequences and SeriesQuestion #411

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

Clue 1 of 3: Arithmetic Progression Properties

For three terms in A.P., represent them as $a - x, a, a + x$ . This simplifies sum and product calculations.

🧠 Full Solution Path

Step 1: Represent AP terms and use sum/product relations✦ Active

Let the three terms of A be $a - d, a, a + d$ . Their sum is $3 a = 36$ , so $a = 12$ . The product is $p = a (a^{2} - d^{2}) = 12 (144 - d^{2})$ . Similarly, for set B, let the terms be $A - D, A, A + D$ . Their sum is $3 A = 36$ , so $A = 12$ . The product is $q = A (A^{2} - D^{2}) = 12 (144 - D^{2})$ . We are given $D = d + 3$ and $d > 0$ .

Step 2: Apply Componendo and Dividendo to the given ratio○ Expand

The given relation is $\frac{p + q}{p - q} = \frac{19}{5}$ . Applying componendo and dividendo, we get $\frac{(p + q) + (p - q)}{(p + q) - (p - q)} = \frac{19 + 5}{19 - 5}$ , which simplifies to $\frac{2 p}{2 q} = \frac{24}{14}$ , so $\frac{p}{q} = \frac{12}{7}$ . Substituting the expressions for $p$ and $q$ : $\frac{12 (144 - d^{2})}{12 (144 - (d + 3)^{2})} = \frac{12}{7}$ This simplifies to $\frac{144 - d^{2}}{144 - (d^{2} + 6 d + 9)} = \frac{12}{7} ⟹ \frac{144 - d^{2}}{135 - d^{2} - 6 d} = \frac{12}{7}$ . Cross-multiplying gives $7 (144 - d^{2}) = 12 (135 - d^{2} - 6 d)$ , which expands to $1008 - 7 d^{2} = 1620 - 12 d^{2} - 72 d$ . Rearranging terms, we get the quadratic equation $5 d^{2} + 72 d - 612 = 0$ .

Step 3: Solve for

d

and calculate

p - q

○ Expand

Using the quadratic formula for $5 d^{2} + 72 d - 612 = 0$ : $d = \frac{- 72 \pm \sqrt{72^{2} - 4 (5) (- 612)}}{2 (5)} = \frac{- 72 \pm \sqrt{5184 + 12240}}{10} = \frac{- 72 \pm \sqrt{17424}}{10}$ . Since $\sqrt{17424} = 132$ , we have $d = \frac{- 72 \pm 132}{10}$ . As $d > 0$ , we take the positive root: $d = \frac{- 72 + 132}{10} = \frac{60}{10} = 6$ . Now, we need to find $p - q$ . We know $p = 12 (144 - d^{2})$ and $q = 12 (144 - D^{2})$ . So, $p - q = 12 (144 - d^{2}) - 12 (144 - D^{2}) = 12 (D^{2} - d^{2})$ . Substitute $D = d + 3$ : $p - q = 12 ((d + 3)^{2} - d^{2}) = 12 (d^{2} + 6 d + 9 - d^{2}) = 12 (6 d + 9)$ . Substitute $d = 6$ : $p - q = 12 (6 (6) + 9) = 12 (36 + 9) = 12 (45) = 540$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 6 – JEE-MAIN 2025

Clue 1 of 3: Arithmetic Progression Properties

🧠 Full Solution Path