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Maths Question 6 – JEE-MAIN 2025

Let x1,x2,x3,x4 be in a geometric progression. If 2,7,9,5 are subtracted respectively from x1,x2,x3,x4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4) is:

Represent the geometric progression (GP) terms using a first term a and common ratio r, and then form the arithmetic progression (AP) terms by subtracting the given constants.

Step 1: Set up GP and AP terms✦ Active

Let the geometric progression be x1=a,x2=ar,x3=ar2,x4=ar3. The terms after subtraction are x1=a2, x2=ar7, x3=ar29, x4=ar35. These terms are in an arithmetic progression.

Step 2: Apply AP conditions and solve for a and r○ Expand

Using the AP property x2x1=x3x2:

(ar7)(a2)=(ar29)(ar7) a(r1)5=ar(r1)2 a(r1)2=3(1)

Using the AP property x3x2=x4x3:

(ar29)(ar7)=(ar35)(ar29) ar2ar2=ar3ar2+4 ar(r1)2=6(2)

Dividing equation (2) by (1) (assuming r1):

ar(r1)2a(r1)2=63r=2

Substitute r=2 into equation (1):

a(21)2=3a=3a=3

Thus, a=3 and r=2. The GP terms are x1=3,x2=6,x3=12,x4=24.

💡 Teacher's Secret Hint

Remember to check the case r=1 to ensure the division is valid. In this case, r=1 leads to a contradiction, so it's safe to divide.

Step 3: Calculate the required value○ Expand

The product of the GP terms is x1x2x3x4=aarar2ar3=a4r6.

x1x2x3x4=(3)4(2)6=81×64=5184

The value to be found is 124(x1x2x3x4):

124(5184)=216
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