Logo
Jeeify
MCQMediumJEE 2025Geometric Progression (GP)

JEE Mathematics 2025 Question with Solution

Let x1,x2,x3,x4x_1, x_2, x_3, x_4 be in a geometric progression. If 22, 77, 99, 55 are subtracted respectively from x1,x2,x3,x4x_1, x_2, x_3, x_4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4)\frac{1}{24}(x_1 x_2 x_3 x_4) is:

  • A

    7272

  • B

    1818

  • C

    3636

  • D

    216216

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: x1,x2,x3,x4x_1, x_2, x_3, x_4 are in G.P.

After subtracting 2,7,9,52, 7, 9, 5 respectively, the numbers x1−2,x2−7,x3−9,x4−5x_1-2, x_2-7, x_3-9, x_4-5 are in A.P.

Find: 124(x1x2x3x4)\frac{1}{24}(x_1 x_2 x_3 x_4)

Let

x1=a,x2=ar,x3=ar2,x4=ar3x_1=a, \quad x_2=ar, \quad x_3=ar^2, \quad x_4=ar^3

Since the modified terms are in A.P.,

(ar−7)−(a−2)=(ar2−9)−(ar−7)(ar-7)-(a-2)=(ar^2-9)-(ar-7)

So,

a(r−1)−5=ar(r−1)−2a(r-1)-5=ar(r-1)-2

Hence,

a(r−1)2=−3...(i)a(r-1)^2=-3 \qquad \text{...(i)}

Also, using the next A.P. relation,

(ar−7)−(a−2)=(ar3−5)−(ar2−9)(ar-7)-(a-2)=(ar^3-5)-(ar^2-9)

So,

a(r−1)−5=ar2(r−1)+4a(r-1)-5=ar^2(r-1)+4

Hence,

a(r−1)(r2−1)=−9...(ii)a(r-1)(r^2-1)=-9 \qquad \text{...(ii)}

Dividing (ii) by (i),

a(r−1)(r2−1)a(r−1)2=−9−3\frac{a(r-1)(r^2-1)}{a(r-1)^2}=\frac{-9}{-3}

Therefore,

r+1=3r+1=3

so,

r=2r=2

Substituting r=2r=2 in (i),

a(1)2=−3a(1)^2=-3

Thus,

a=−3a=-3

Therefore,

x1=−3,x2=−6,x3=−12,x4=−24x_1=-3, \quad x_2=-6, \quad x_3=-12, \quad x_4=-24

Now,

124(x1x2x3x4)=124(−3)(−6)(−12)(−24)=216\frac{1}{24}(x_1 x_2 x_3 x_4)=\frac{1}{24}(-3)(-6)(-12)(-24)=216

Therefore, the correct option is D.

Using consecutive common differences

Given: x1,x2,x3,x4x_1, x_2, x_3, x_4 are in G.P., and x1−2,x2−7,x3−9,x4−5x_1-2, x_2-7, x_3-9, x_4-5 are in A.P.

Find: 124(x1x2x3x4)\frac{1}{24}(x_1 x_2 x_3 x_4)

Write the G.P. terms as

x2=x1r,x3=x1r2,x4=x1r3x_2=x_1r, \quad x_3=x_1r^2, \quad x_4=x_1r^3

For four numbers in A.P., consecutive differences are equal:

(x2−7)−(x1−2)=(x3−9)−(x2−7)=(x4−5)−(x3−9)(x_2-7)-(x_1-2)=(x_3-9)-(x_2-7)=(x_4-5)-(x_3-9)

This gives

x2−x1−5=x3−x2−2=x4−x3+4x_2-x_1-5=x_3-x_2-2=x_4-x_3+4

Using the first two expressions,

x1r−x1−5=x1r2−x1r−2x_1r-x_1-5=x_1r^2-x_1r-2

which simplifies to

x1(r−1)2=−3x_1(r-1)^2=-3

Using the first and third expressions,

x1r−x1−5=x1r3−x1r2+4x_1r-x_1-5=x_1r^3-x_1r^2+4

which simplifies to

x1(r−1)(r2−1)=−9x_1(r-1)(r^2-1)=-9

Dividing the second equation by the first,

x1(r−1)(r2−1)x1(r−1)2=3\frac{x_1(r-1)(r^2-1)}{x_1(r-1)^2}=3

So,

r2−1r−1=3\frac{r^2-1}{r-1}=3

that is,

r+1=3⇒r=2r+1=3 \Rightarrow r=2

Then

x1(2−1)2=−3⇒x1=−3x_1(2-1)^2=-3 \Rightarrow x_1=-3

Hence the G.P. is

−3,−6,−12,−24-3, -6, -12, -24

Therefore,

124(x1x2x3x4)=124(−3)(−6)(−12)(−24)=216\frac{1}{24}(x_1x_2x_3x_4)=\frac{1}{24}(-3)(-6)(-12)(-24)=216

So the required value is 216216 and the correct option is D.

Common mistakes

  • Assuming the A.P. condition means only one pair of differences is equal. For four terms in A.P., the common difference must be consistent across the sequence. Use equal-difference relations carefully before substituting the G.P. terms.

  • Making a sign error while simplifying the subtracted terms, especially with −5-5, −2-2, and +4+4. These constants come from expressions like (ar−7)−(a−2)(ar-7)-(a-2) and must be expanded carefully.

  • Using the incorrect value of the first term from the flawed solution step. The correct substitution from a(r−1)2=−3a(r-1)^2=-3 with r=2r=2 gives a=−3a=-3, not a=3a=3. The final product remains positive because all four terms multiply to a positive number.

Practice more Geometric Progression (GP) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Geometric Progression (GP)

Related questions