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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Sequences and Series

Let x_1, x_2, x_3, x_4 be in a geometric progression. 2,7,9,5 are...

Let x1,x2,x3,x4\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4 be in a geometric progression. 2,7,9,52,7,9,5 are subtracted respectively from x1,x2,x3x_1, x_2, x_3 x4thentheresultingnumbersareinanarithmeticprogression.Thenthevalueofx_4 then the resulting numbers are in an arithmetic progression. Then the value of\frac{1}{24}(x1x2x3x4)\left(x_1 x_2 x_3 x_4\right) is :

Options

  1. A72
  2. B18
  3. C36
  4. D216

Correct answer

D. 216

Step-by-step solution

x1,x2,x3,x4→ G.P. \mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4 \rightarrow \text{ G.P. }
Let a, ar, ar2,ar3→\mathrm{ar}^2, \mathrm{ar}^3 \rightarrow G.P.
Now a−2,ar−7,a2−9,ar3−5→a-2, a r-7, a^2-9, \mathrm{ar}^3-5 \rightarrow A.P.
2(ar−7)=a−2+ar2−92(\mathrm{ar}-7)=\mathrm{a}-2+\mathrm{ar}^2-9....(i)
2(ar⁡2−9)=ar−7+ar3−52\left(\operatorname{ar}^2-9\right)=\mathrm{ar}-7+\mathrm{ar}^3-5....(ii)
 Solving r=2,a=−3∴ Product =x1,x2,x3,x4=a4r6=81×64\begin{aligned} & \text{ Solving } \mathrm{r}=2, \mathrm{a}=-3 \\ & \therefore \text{ Product }=\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4=\mathrm{a}^4 \mathrm{r}^6=81 \times 64\end{aligned}
 Solving r=2,a=−3∴ Product =x1,x2,x3,x4=a4r6=81×64\begin{aligned} & \text{ Solving } \mathrm{r}=2, \mathrm{a}=-3 \\ & \therefore \text{ Product }=\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4=\mathrm{a}^4 \mathrm{r}^6=81 \times 64\end{aligned}
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