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}\left(x_1 x_2 x_3 x_4\right) is:

Question

Accepted Answer

Answer: B. 216 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

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

Answer

Explanation

More JEE Math questions