Let x_1, x_2, x_3, x_4 be in a geometric progression.

Question

Let $x_1, x_2, x_3, x_4$ be in a geometric progression. If $2,7,9,5$ are subtracted respectively from $x_1, x_2, x_3, x_4$

Accepted Answer

Given the sequence $x_1, x_2, x_3, x_4$ in geometric progression:

$x_1 = a$

$x_2 = ar$

$x_3 = ar^2$

$x_4 = ar^3$

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:

$a - 2$

$ar - 7$

$ar^2 - 9$

$ar^3 - 5$

For these to form an arithmetic progression, the common differences must be equal, so:

$ (ar - 7) - (a - 2) = (ar^2 - 9) - (ar - 7) $

Simplifying gives:

$    a(r - 1) - 5 = ar(r - 1) - 2    $

$    a(r - 1)(r - 1) = -3 \quad (i)    $

$ (ar - 7) - (a - 2) = (ar^3 - 5) - (ar^2 - 9) $

Simplifying gives:

$    a(r - 1) - 5 = ar^2(r - 1) + 4    $

$    a(r - 1)(r^2 - 1) = -9 \quad (ii)    $

Using the ratio of equations (ii) and (i):

$ \frac{a(r - 1)(r^2 - 1)}{a(r - 1)(r - 1)} = \frac{-9}{-3} $

$ r + 1 = 3 \implies r = 2 $

Plugging back into equation (i):

$ a(1)(1) = -3 \implies a = -3 $

So, the sequence $x_1, x_2, x_3, x_4$ is:

$x_1 = -3$

$x_2 = -6$

$x_3 = -12$

$x_4 = -24$

The expression for $\frac{1}{24}(x_1 \cdot x_2 \cdot x_3 \cdot x_4)$ is:

$ \frac{1}{24}((-3) \cdot (-6) \cdot (-12) \cdot (-24)) = 216 $

Solutions · 1