JEE Main 2025 Apr 7 Shift 1, Mathematics Q13: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 7, Shift 1Mathematics

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. If $2, 7, 9, 5$ are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is :

A
$72$
B
$18$
C
$36$
D
$216$

Solution

Let the GP be $a, a r, a r^{2}, a r^{3}$ . After the subtractions, the AP condition gives two equations by equating consecutive differences.

First AP condition (middle-term identity applied to first three AP terms):

$2 (a r - 7) = (a - 2) + (a r^{2} - 9)$

$\Rightarrow 2 a r - 14 = a + a r^{2} - 11 \Rightarrow a r^{2} - 2 a r + a = - 3 \Rightarrow a (r - 1)^{2} = - 3 \dots (i)$

Second AP condition (middle-term identity for last three):

$2 (a r^{2} - 9) = (a r - 7) + (a r^{3} - 5)$

$\Rightarrow 2 a r^{2} - 18 = a r + a r^{3} - 12 \Rightarrow a r (r - 1)^{2} = - 6 \dots (ii)$

Divide (ii) by (i): $r = 2$ .

Substitute back: $a (2 - 1)^{2} = - 3 \Rightarrow a = - 3$ .

Product: $x_{1} x_{2} x_{3} x_{4} = a \cdot a r \cdot a r^{2} \cdot a r^{3} = a^{4} r^{6} = 81 \times 64 = 5184$ .

$\frac{1}{24} (x_{1} x_{2} x_{3} x_{4}) = \frac{5184}{24} = 216$

$∴$ Option (4).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q13: Arithmetic, Geometric And Harmonic Progressions

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