Let x_1, x_2, x_3, x_4 be in a geometric prog | vedclass

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(D) Let the terms of the geometric progression be $a, ar, ar^2, ar^3$.Given that $a-2, ar-7, ar^2-9, ar^3-5$ are in an arithmetic progression.For an $

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$216$

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(D) Let the terms of the geometric progression be $a, ar, ar^2, ar^3$.Given that $a-2, ar-7, ar^2-9, ar^3-5$ are in an arithmetic progression.For an $A$.$P$. with terms $A, B, C, D$,we have $2B = A+C$ and $2C = B+D$.$2(ar-7) = (a-2) + (ar^2-9) \implies 2ar - 14 = a + ar^2 - 11 \implies a(r^2 - 2r + 1) = -3 \implies a(r-1)^2 = -3 \dots(i)$$2(ar^2-9) = (ar-7) + (ar^3-5) \implies 2ar^2 - 18 = ar + ar^3 - 12 \implies ar(r^2 - 2r + 1) = -6 \implies ar(r-1)^2 = -6 \dots(ii)$Dividing $(ii)$ by $(i)$,we get $r = \frac{-6}{-3} = 2$.Substituting $r=2$ in $(i)$,$a(2-1)^2 = -3 \implies a = -3$.The terms are $x_1 = -3, x_2 = -6, x_3 = -12, x_4 = -24$.The product $x_1 x_2 x_3 x_4 = (-3)(-6)(-12)(-24) = 5184$.The value of $\frac{1}{24}(x_1 x_2 x_3 x_4) = \frac{5184}{24} = 216$.

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

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