JEE Main 2025 Sequence and series with solution

Question :

Consider two set 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 difference 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

Options

(a) 600

(b) 450

(c) 630

(d) 540

JEE Main 2025 Sequence and series with solution

Answer :

Set – A

Sum of elements = a + d + a + a – d = 36

3a = 36

a = 12

Product of elements : ( a + d )( a)( a – d) = p

a ( a²– d²) = p

12( 144 – d²) = p ————————1

Set – B

Sum of elements = b – B + b + b – B = 36

3b = 36

b = 12

Product of elements = b( b² – D² ) = q

12( 144 – D²) = q—————————- 2

Given

$\frac{P+q}{p-q}$ = $\frac{19}{5}$ $\rightarrow$ Componendo and Dividendo

$\frac{2p}{2q}$ = $\frac{24}{14}$

$\frac{p}{q}$ = $\frac{12}{7}$

divide equation 1 and 2

$\frac{12(144 – d^2) }{12(144 – D^2)}$ = $\frac{p}{q}$

$\frac{144 – d^2}{ 144 – D^2}$ = $\frac{12}{7}$

$\frac{144 – d^2}{144 – (d+3)^2 }$ = $\frac{12}{7}$

7(144 – d²) = 12( 144 – d² – 9 – 6d)

-5(144) + 5d² + 72d + 108 = 0

5d² + 72d – 612 = 0

5d² – 30d + 102d – 612 = 0

5d( d – 6 ) + 102 (d – 6 ) = 0

d = 6

substitude d = 6 in 1st equation

12( 144 – d²) = p

12( 144 – 36 ) = p

substitude d = 6 2nd equation

12 ( 144 – D²) = q

12 ( 144 – 81 ) = q

p – q = 12 (144 – 36 ) – 12 ( 144 – 81)

12 ( 108 – 63)

= 12(45)

p – q = 540

Correct answer : Option (d)

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