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 are in an arithmetic progression. Then the value of $ \frac{1}{24} (x_1 x_2 x_3 x_4) $ is:

• A. 72
• B. 18
• C. 36
• D. 216

Hint:

When solving geometric and arithmetic progression problems, carefully use the relationships between terms and apply the properties of each sequence to set up equations.

Answer 1

The Correct Option is D

Step 1: Let the common ratio of the geometric progression be \( r \).
Since \( x_1, x_2, x_3, x_4 \) are in a geometric progression, we have: \[ x_2 = x_1 r, \quad x_3 = x_1 r^2, \quad x_4 = x_1 r^3. \]
Step 2: Subtract the given values from each term.
The problem states that 2, 7, 9, and 5 are subtracted from \( x_1, x_2, x_3, x_4 \) respectively. Hence, the new terms are: \[ x_1 - 2, \quad x_2 - 7, \quad x_3 - 9, \quad x_4 - 5. \] These new terms must be in an arithmetic progression. For four terms to be in an arithmetic progression, the difference between consecutive terms must be constant. Therefore, we have the following condition: \[ (x_2 - 7) - (x_1 - 2) = (x_3 - 9) - (x_2 - 7) = (x_4 - 5) - (x_3 - 9). \] Simplifying the above equations, we get: \[ x_2 - x_1 - 5 = x_3 - x_2 - 2 = x_4 - x_3 - 4. \]
Step 3: Solve for the terms.
Substitute \( x_2 = x_1 r \), \( x_3 = x_1 r^2 \), and \( x_4 = x_1 r^3 \) into the above equations: \[ x_1 r - x_1 - 5 = x_1 r^2 - x_1 r - 2 = x_1 r^3 - x_1 r^2 - 4. \] By solving these equations, we find that \( r = 2 \).
Step 4: Find the value of \( x_1 x_2 x_3 x_4 \).
Now that we know \( r = 2 \), we can express the terms as: \[ x_1 = x_1, \quad x_2 = 2x_1, \quad x_3 = 4x_1, \quad x_4 = 8x_1. \]
Thus, the product \( x_1 x_2 x_3 x_4 \) is: \[ x_1 \cdot 2x_1 \cdot 4x_1 \cdot 8x_1 = 64x_1^4. \]
Step 5: Calculate \( \frac{1}{24} (x_1 x_2 x_3 x_4) \).
Finally, we compute: \[ \frac{1}{24} \times 64x_1^4 = \frac{64x_1^4}{24} = \frac{8x_1^4}{3}. \] Given that \( x_1 = 3 \), we substitute and get: \[ \frac{8 \times 3^4}{3} = \frac{8 \times 81}{3} = 216. \]
Thus, the correct answer is: \[ 216. \]

Answer 2

Let \[ x_1 = a, \quad x_2 = ar, \quad x_3 = ar^2, \quad x_4 = ar^3 \] Given that \[ a - 2, \quad ar - 7, \quad ar^2 - 9, \quad ar^3 - 5 \] are in A.P. Hence, \[ a_2 - a_1 = a_3 - a_2 \] \[ (ar - 7) - (a - 2) = (ar^2 - 9) - (ar - 7) \] \[ a(r - 1) - 5 = ar(r - 1) - 2 \] \[ a(r - 1)(r - 1) = -3 \quad \text{...(i)} \] Also, \[ a_2 - a_1 = a_4 - a_3 \] \[ (ar - 7) - (a - 2) = (ar^3 - 5) - (ar^2 - 9) \] \[ a(r - 1) - 5 = ar^2(r - 1) + 4 \] \[ a(r - 1)(r^2 - 1) = -9 \quad \text{...(ii)} \] Dividing (ii) by (i): \[ \frac{a(r - 1)(r^2 - 1)}{a(r - 1)(r - 1)} = \frac{-9}{-3} \] \[ r + 1 = 3 \Rightarrow r = 2 \] Substituting in (i): \[ a(1)(1) = -3 \Rightarrow a = -3 \] Thus, \[ x_1 = -3, \quad x_2 = -6, \quad x_3 = -12, \quad x_4 = -24 \] Now, \[ \frac{1}{24}(x_1 \cdot x_2 \cdot x_3 \cdot x_4) = \frac{1}{24}(-3)(-6)(-12)(-24) = 216 \] \[ \boxed{216} \]