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 $x_1,x_2,x_3,x_4$, the resulting numbers are in an arithmetic progression. Then the value of $\dfrac{1}{24}(x_1x_2x_3x_4)$ is:

Accepted Answer

Let $x_1=a, x_2=ar, x_3=ar^2, x_4=ar^3$. The A.P. condition gives:
$$a(r-1)(r-1) = -3 \quad(i), \qquad a(r-1)(r^2-1) = -9 \quad(ii).$$

Dividing (ii) by (i): $r+1 = 3 \Rightarrow r=2$. Then from (i): $a=-3$.

So $x_1=-3, x_2=-6, x_3=-12, x_4=-24$, and
$$\frac{1}{24}x_1x_2x_3x_4 = \frac{1}{24}(-3)(-6)(-12)(-24) = 216.$$

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