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 $\frac{1}{24}(x_1x_2x_3x_4)$ is:

• A. 216
• B. 108
• C. 432
• D. 5184

Answer 1

Answer: (A)

216

Answer 2

Let the geometric progression be $x_1 = a$, $x_2 = ar$, $x_3 = ar^2$, and $x_4 = ar^3$. When 2, 7, 9, and 5 are subtracted from $x_1, x_2, x_3, x_4$ respectively, the resulting numbers are $a-2$, $ar-7$, $ar^2-9$, and $ar^3-5$. These numbers form an arithmetic progression. Therefore, the difference between consecutive terms is constant: $(ar-7) - (a-2) = (ar^2-9) - (ar-7)$ $ar - a - 5 = ar^2 - ar - 2$ $ar^2 - 2ar + a + 3 = 0$ $a(r^2 - 2r + 1) + 3 = 0$ $a(r-1)^2 = -3 \quad (*)$

Also, $(ar^2-9) - (ar-7) = (ar^3-5) - (ar^2-9)$ $ar^2 - ar - 2 = ar^3 - ar^2 + 4$ $ar^3 - 2ar^2 + ar + 6 = 0$ $ar(r^2 - 2r + 1) + 6 = 0$ $ar(r-1)^2 = -6 \quad (**)$

Substitute the value of $a(r-1)^2$ from $(*)$ into $(**)$: $r \cdot [a(r-1)^2] = -6$ $r \cdot (-3) = -6$ $-3r = -6$ $r = 2$

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

The terms of the geometric progression are $x_1 = -3$, $x_2 = (-3)(2) = -6$, $x_3 = (-3)(2^2) = -12$, $x_4 = (-3)(2^3) = -24$.

The product $x_1x_2x_3x_4 = a \cdot ar \cdot ar^2 \cdot ar^3 = a^4r^6$. $x_1x_2x_3x_4 = (-3)^4 (2)^6 = 81 \times 64 = 5184$.

The value to be found is $\frac{1}{24}(x_1x_2x_3x_4)$: $\frac{1}{24}(5184) = \frac{5184}{24} = 216$.