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 $\frac{1}{24}\left(x_1 x_2 x_3 x_4\right)$ is:

<p>Given the sequence $x_1, x_2, x_3, x_4$ in geometric progression:</p> <p><p>$x_1 = a$</p></p> <p><p>$x_2 = ar$</p></p> <p><p>$x_3 = ar^2$</p></p> <p><p>$x_4 = ar^3$</p></p> <p>When you subtract 2, 7, 9, and 5 from $x_1, x_2, x_3, x_4$ respectively, the sequence becomes an arithmetic progression. Thus, the new sequence is:</p> <p><p>$a - 2$</p></p> <p><p>$ar - 7$</p></p> <p><p>$ar^2 - 9$</p></p> <p><p>$ar^3 - 5$</p></p> <p>For these to form an arithmetic progression, the common differences must be equal, so:</p> <p><p>$ (ar - 7) - (a - 2) = (ar^2 - 9) - (ar - 7) $</p> <p>Simplifying gives:</p> <p>$ a(r - 1) - 5 = ar(r - 1) - 2 $</p> <p>$ a(r - 1)(r - 1) = -3 \quad (i) $</p></p> <p><p>$ (ar - 7) - (a - 2) = (ar^3 - 5) - (ar^2 - 9) $</p> <p>Simplifying gives:</p> <p>$ a(r - 1) - 5 = ar^2(r - 1) + 4 $</p> <p>$ a(r - 1)(r^2 - 1) = -9 \quad (ii) $</p></p> <p>Using the ratio of equations (ii) and (i):</p> <p>$ \frac{a(r - 1)(r^2 - 1)}{a(r - 1)(r - 1)} = \frac{-9}{-3} $</p> <p>$ r + 1 = 3 \implies r = 2 $</p> <p>Plugging back into equation (i):</p> <p>$ a(1)(1) = -3 \implies a = -3 $</p> <p>So, the sequence $x_1, x_2, x_3, x_4$ is:</p> <p><p>$x_1 = -3$</p></p> <p><p>$x_2 = -6$</p></p> <p><p>$x_3 = -12$</p></p> <p><p>$x_4 = -24$</p></p> <p>The expression for $\frac{1}{24}(x_1 \cdot x_2 \cdot x_3 \cdot x_4)$ is:</p> <p>$ \frac{1}{24}((-3) \cdot (-6) \cdot (-12) \cdot (-24)) = 216 $</p>