Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The terms are $a, ar, ar^2, ar^3, \dots$.
• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant, known as the common difference ($D$). The terms are $A, A+D, A+2D, A+3D, \dots$. The defining property is $T_{n+1} - T_n = D$ for all $n$.
• Product of GP terms: For a GP $x_1, x_2, x_3, x_4$, where $x_1=a$ and common ratio is $r$, the product is $x_1 x_2 x_3 x_4 = a \cdot ar \cdot ar^2 \cdot ar^3 = a^4 r^6$.

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Step-by-Step Solution

Step 1: Represent the terms of the GP. Let the four terms in geometric progression be $x_1, x_2, x_3, x_4$. We can represent them in terms of the first term ($a$) and the common ratio ($r$) as: $x_1 = a$ $x_2 = ar$ $x_3 = ar^2$ $x_4 = ar^3$

Step 2: Represent the terms of the AP. When $2, 7, 9, 5$ are subtracted respectively from $x_1, x_2, x_3, x_4$, the resulting numbers are in an arithmetic progression. Let these new terms be $y_1, y_2, y_3, y_4$. $y_1 = x_1 - 2 = a - 2$ $y_2 = x_2 - 7 = ar - 7$ $y_3 = x_3 - 9 = ar^2 - 9$ $y_4 = x_4 - 5 = ar^3 - 5$

Step 3: Use the property of AP to form equations. Since $y_1, y_2, y_3, y_4$ are in an arithmetic progression, the difference between consecutive terms is constant. Therefore, we have: $y_2 - y_1 = y_3 - y_2$ (Equation A) $y_3 - y_2 = y_4 - y_3$ (Equation B)

Let's substitute the expressions for $y_i$ into these equations.

From Equation A: $(ar - 7) - (a - 2) = (ar^2 - 9) - (ar - 7)$ $ar - 7 - a + 2 = ar^2 - 9 - ar + 7$ $ar - a - 5 = ar^2 - ar - 2$ $a(r - 1) - 5 = ar(r - 1) - 2$ Rearranging terms to group $a(r-1)$: $a(r - 1) - ar(r - 1) = -2 + 5$ $a(r - 1)(1 - r) = 3$ Since $(1 - r) = -(r - 1)$, we have: $a(r - 1)(-(r - 1)) = 3$ $-a(r - 1)^2 = 3$ $a(r - 1)^2 = -3$ (Equation 1)

From Equation B: $(ar^2 - 9) - (ar - 7) = (ar^3 - 5) - (ar^2 - 9)$ $ar^2 - 9 - ar + 7 = ar^3 - 5 - ar^2 + 9$ $ar^2 - ar - 2 = ar^3 - ar^2 + 4$ $ar(r - 1) - 2 = ar^2(r - 1) + 4$ Rearranging terms: $ar(r - 1) - ar^2(r - 1) = 4 + 2$ $ar(r - 1)(1 - r) = 6$ Using $(1 - r) = -(r - 1)$: $ar(r - 1)(-(r - 1)) = 6$ $-ar(r - 1)^2 = 6$ $ar(r - 1)^2 = -6$ (Equation 2)

Step 4: Solve the system of equations for $a$ and $r$. We have the system:

1. $a(r - 1)^2 = -3$
2. $ar(r - 1)^2 = -6$

Divide Equation 2 by Equation 1: $\frac{ar(r - 1)^2}{a(r - 1)^2} = \frac{-6}{-3}$ Assuming $a \neq 0$ and $(r-1)^2 \neq 0$ (which is implied by Equation 1 being non-zero), we can cancel terms: $r = 2$

Substitute $r=2$ into Equation 1: $a(2 - 1)^2 = -3$ $a(1)^2 = -3$ $a = -3$

So, the first term of the GP is $a = -3$ and the common ratio is $r = 2$.

Step 5: Calculate the terms of the GP. $x_1 = a = -3$ $x_2 = ar = (-3)(2) = -6$ $x_3 = ar^2 = (-3)(2^2) = -12$ $x_4 = ar^3 = (-3)(2^3) = -24$

Step 6: Calculate the product $x_1 x_2 x_3 x_4$. The product can be calculated directly or using the formula $a^4 r^6$: $x_1 x_2 x_3 x_4 = (-3)(-6)(-12)(-24)$ $x_1 x_2 x_3 x_4 = (18)(288)$ $x_1 x_2 x_3 x_4 = 5184$

Alternatively, using the formula: $x_1 x_2 x_3 x_4 = a^4 r^6 = (-3)^4 (2)^6 = (81)(64) = 5184$.

Step 7: Calculate the final required value. We need to find $\frac{1}{24}\left(x_1 x_2 x_3 x_4\right)$. $\frac{1}{24}(5184) = \frac{5184}{24}$ To simplify the division: $\frac{5184}{24} = \frac{2592}{12} = \frac{1296}{6} = \frac{648}{3} = 216$.

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Common Mistakes & Tips

• Algebraic Errors: Be meticulous when expanding and simplifying expressions, especially when dealing with negative signs and factored terms like $(r-1)$.
• Checking the AP: After finding $a$ and $r$, it's a good practice to verify that the resulting $y_i$ terms actually form an AP. For $a=-3, r=2$: $y_1 = -3 - 2 = -5$ $y_2 = -6 - 7 = -13$ $y_3 = -12 - 9 = -21$ $y_4 = -24 - 5 = -29$ The differences are $-13 - (-5) = -8$, $-21 - (-13) = -8$, $-29 - (-21) = -8$. This confirms the AP.
• Division by Zero: Ensure that when dividing equations, the divisor is not zero. Equation 1, $a(r-1)^2 = -3$, guarantees that $a \neq 0$ and $r \neq 1$, so division is valid.

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Summary

The problem involves relating terms of a geometric progression to terms of an arithmetic progression. By defining the GP terms as $a, ar, ar^2, ar^3$ and the resulting AP terms as $a-2, ar-7, ar^2-9, ar^3-5$, we used the property of constant common difference in AP to set up two equations involving $a$ and $r$. Solving these equations yielded $a=-3$ and $r=2$. We then calculated the product of the GP terms $x_1 x_2 x_3 x_4$ and divided by 24 to find the final answer.

The final answer is $\boxed{216}$.