JEE Main: Geometric to Arithmetic Progression — Fast Method 💡

 

❓ Question

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 form an arithmetic progression.

Find the value of

$$\frac{1}{24}\left(x_1{x}_2{x}_3{x}_4\right)\mathrm{.}$$

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🖼️ Question Image

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✍️ Short Solution

We’ll express the GP in standard form, apply the AP equal-difference condition, solve for the first term & common ratio, then compute the product — and finally divide by 24.

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🔹 Step 1 — Assume the GP

Let

$$x_1=a,\quad x_2=ar,\quad x_3=ar^2,\quad x_4=ar^3$$

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🔹 Step 2 — Form the AP after subtraction

Given that

$$a-2,\quad ar-7,\quad ar^2-9,\quad ar^3-5$$

are in AP, so:

$$(ar-7)-(a-2) = (ar^2-9)-(ar-7)$$

and

$$(ar^2-9)-(ar-7) = (ar^3-5)-(ar^2-9)$$

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🔹 Step 3 — Simplify the first equality

$$a(r-1)-5 = a(r^2-r)-2$$

Bring like terms together:

$$a(2r-1-r^2) = 3$$

Note:

$$2r-1-r^2 = -(r-1)^2$$

So:

$$-a(r-1)^2 = 3 \quad\Rightarrow\quad a(r-1)^2 = -3 \quad (1)$$

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🔹 Step 4 — Simplify the second equality

$$a(r^2-r)-2 = a(r^3-r^2)+4$$

This becomes:

$$a(2r^2-r-r^3) = 6$$

But:

$$2r^2-r-r^3 = -r(r-1)^2$$

So:

$$-a r(r-1)^2 = 6 \quad\Rightarrow\quad a r(r-1)^2 = -6 \quad (2)$$

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🔹 Step 5 — Divide (2) by (1)

$$\frac{ar(r-1)^2}{a(r-1)^2} = \frac{-6}{-3}$$
$$r = 2$$

Substitute back in (1):

$$a(2-1)^2 = -3 \Rightarrow a = -3$$

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🔹 Step 6 — Write the GP terms

$$x_1=-3,\quad x_2=-6,\quad x_3=-12,\quad x_4=-24$$

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🔹 Step 7 — Verify the AP condition

Subtracting 2,7,9,5 gives:

$$-5,\ -13,\ -21,\ -29$$

Common difference = $-8$ ✔ AP confirmed

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🔹 Step 8 — Compute the product

$$x_1x_2x_3x_4 = (-3)(-6)(-12)(-24) = 5184$$

So,

$$\frac{1}{24}(x_1x_2x_3x_4)=\frac{5184}{24}=216$$

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✅ Final Answer

$$\boxed{216}$$