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Sequences and Series question

2025 · 7 Apr · Shift 1 · Q45
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Sequences and Series question

2025 · 7 Apr · Shift 1 · Q45

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let x1,x2,x3,x4x_1, x_2, x_3, x_4x1​,x2​,x3​,x4​ be in a geometric progression. If 2,7,9,52,7,9,52,7,9,5 are subtracted respectively from x1,x2,x3,x4x_1, x_2, x_3, x_4x1​,x2​,x3​,x4​, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4)\frac{1}{24}\left(x_1 x_2 x_3 x_4\right)241​(x1​x2​x3​x4​) is:
  1. A
    18
  2. B
    216
  3. C
    36
  4. D
    72
View written solutionFree

Correct answer: B

  1. Let the four terms in G.P. be x1=a, x2=ar, x3=ar2, x4=ar3.x_1=a,\, x_2=ar,\, x_3=ar^2,\, x_4=ar^3.x1​=a,x2​=ar,x3​=ar2,x4​=ar3.

  2. Given condition after subtraction

    After subtracting 2,7,9,52,7,9,52,7,9,5 respectively, we get: a−2,ar−7,ar2−9,ar3−5a-2,\quad ar-7,\quad ar^2-9,\quad ar^3-5a−2,ar−7,ar2−9,ar3−5 and these are in A.P.

    Hence consecutive differences are equal: (ar−7)−(a−2)=(ar2−9)−(ar−7)=(ar3−5)−(ar2−9).(ar-7)-(a-2)=(ar^2-9)-(ar-7)=(ar^3-5)-(ar^2-9).(ar−7)−(a−2)=(ar2−9)−(ar−7)=(ar3−5)−(ar2−9).

  3. Form the equations

    First equality: ar−a−5=ar2−ar−2ar-a-5=ar^2-ar-2ar−a−5=ar2−ar−2 a(2r−r2−1)=3a(2r-r^2-1)=3a(2r−r2−1)=3 a[−(r−1)2]=−3a[-(r-1)^2]=-3a[−(r−1)2]=−3 a(r−1)2=−3 (i)a(r-1)^2=-3 \,\text{(i)}a(r−1)2=−3(i)

    Second equality: ar2−ar−2=ar3−ar2+4ar^2-ar-2=ar^3-ar^2+4ar2−ar−2=ar3−ar2+4 a(2r2−r−r3)=−6a(2r^2-r-r^3)= -6a(2r2−r−r3)=−6 ar(2r−1−r2)=−6ar(2r-1-r^2)=-6ar(2r−1−r2)=−6 ar[−(r−1)2]=−6ar[-(r-1)^2]=-6ar[−(r−1)2]=−6 ar(r−1)2=6 (ii)ar(r-1)^2=6 \,\text{(ii)}ar(r−1)2=6(ii)

  4. Divide (ii) by (i) ar(r−1)2a(r−1)2=6−3\frac{ar(r-1)^2}{a(r-1)^2}=\frac{6}{-3}a(r−1)2ar(r−1)2​=−36​ r=−2.r=-2.r=−2.

  5. Find aaa using (i) a(r−1)2=−3a(r-1)^2=-3a(r−1)2=−3 a(−3)2=−3a(-3)^2=-3a(−3)2=−3 9a=−39a=-39a=−3 a=−13.a=-\frac13.a=−31​.

  6. Now find the product x1x2x3x4x_1x_2x_3x_4x1​x2​x3​x4​

    Since x1x2x3x4=a⋅ar⋅ar2⋅ar3=a4r6,x_1x_2x_3x_4=a\cdot ar\cdot ar^2\cdot ar^3=a^4r^6,x1​x2​x3​x4​=a⋅ar⋅ar2⋅ar3=a4r6, substitute a=−13a=-\frac13a=−31​, r=−2r=-2r=−2: x1x2x3x4=(134)(26)=6481.x_1x_2x_3x_4=\left(\frac{1}{3^4}\right)(2^6)=\frac{64}{81}.x1​x2​x3​x4​=(341​)(26)=8164​.

    This seems inconsistent with the options, so it is better to directly compute the terms: x1=−13, x2=23, x3=−43, x4=83.x_1=-\frac13,\, x_2=\frac23,\, x_3=-\frac43,\, x_4=\frac83.x1​=−31​,x2​=32​,x3​=−34​,x4​=38​. Then

    \quad x_2-7=-\frac{19}{3}, \quad x_3-9=-\frac{31}{3}, \quad x_4-5=-\frac73,$$ which are **not** in A.P. So there is an algebraic sign issue above; let us solve carefully using A.P. condition.

  7. Correct A.P. condition using middle-term property

    For four numbers in A.P.: 2(ar−7)=(a−2)+(ar2−9)2(ar-7)=(a-2)+(ar^2-9)2(ar−7)=(a−2)+(ar2−9) 2ar−14=a−2+ar2−92ar-14=a-2+ar^2-92ar−14=a−2+ar2−9 2ar−a−ar2=32ar-a-ar^2=32ar−a−ar2=3 a(2r−1−r2)=3a(2r-1-r^2)=3a(2r−1−r2)=3 −a(r−1)2=3-a(r-1)^2=3−a(r−1)2=3 a(r−1)2=−3(1)a(r-1)^2=-3 \quad \text{(1)}a(r−1)2=−3(1)

    Also, 2(ar2−9)=(ar−7)+(ar3−5)2(ar^2-9)=(ar-7)+(ar^3-5)2(ar2−9)=(ar−7)+(ar3−5) 2ar2−18=ar−7+ar3−52ar^2-18=ar-7+ar^3-52ar2−18=ar−7+ar3−5 2ar2−ar−ar3=62ar^2-ar-ar^3=62ar2−ar−ar3=6 ar(2r−1−r2)=6ar(2r-1-r^2)=6ar(2r−1−r2)=6 −ar(r−1)2=6-ar(r-1)^2=6−ar(r−1)2=6 ar(r−1)2=−6(2)ar(r-1)^2=-6 \quad \text{(2)}ar(r−1)2=−6(2)

  8. Divide (2) by (1) r=−6−3=2.r=\frac{-6}{-3}=2.r=−3−6​=2.

  9. Find aaa a(r−1)2=−3a(r-1)^2=-3a(r−1)2=−3 a(1)2=−3a(1)^2=-3a(1)2=−3 a=−3.a=-3.a=−3.

    Therefore the G.P. is: x1=−3,x2=−6,x3=−12,x4=−24.x_1=-3,\quad x_2=-6,\quad x_3=-12,\quad x_4=-24.x1​=−3,x2​=−6,x3​=−12,x4​=−24.

  10. Check the A.P. condition After subtraction:

    \quad -6-7=-13, \quad -12-9=-21, \quad -24-5=-29.$$ These are in A.P. with common difference $-8$.

  11. Compute the required value x1x2x3x4=(−3)(−6)(−12)(−24)=5184.x_1x_2x_3x_4=(-3)(-6)(-12)(-24)=5184.x1​x2​x3​x4​=(−3)(−6)(−12)(−24)=5184.

    Hence, 124(x1x2x3x4)=518424=216.\frac{1}{24}(x_1x_2x_3x_4)=\frac{5184}{24}=216.241​(x1​x2​x3​x4​)=245184​=216.

  12. Final answer The correct option is: 216\boxed{216}216​ i.e. Option B.

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