Let be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from , then the resulting numbers are in an arithmetic progression. Then the value of is : [2025]

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Solution

Q.
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 : [2025]

1 36

2 216

3 72

4 18

Ans.
(2)

Given, $x_{1}, x_{2}, x_{3}, x_{4}$ be in a G.P.

$x_{1} = a, x_{2} = a r, x_{3} = a r^{2}, x_{4} = a r^{3}$

According to question

$a - 2, a r - 7, a r^{2} - 9, a r^{3} - 5$ are in A.P.

So, $2 (a r - 7) = a - 2 + a r^{2} - 9$ ... (i)

$2 (a r^{2} - 9) = a r - 7 + a r^{3} - 5$ ... (ii)

Solving (i) and (ii), we get r = 2, a = – 3

Product $= x_{1} x_{2} x_{3} x_{4} = a^{4} r^{6} = 81 \times 64 = 5184$

$∴$ The value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4}) = \frac{(5184)}{24} = 216$ .

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