Let mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4$\mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3, \mathbf{x}_4$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4$\mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3, \mathbf{x}_4$ then the resulting numbers are in an arithmetic progression. Then the value of frac124 (mathbfx_1 mathbfx_2 mathbfx_3 mathbfx_4)$\frac{1}{24} (\mathbf{x}_1 \mathbf{x}_2 \mathbf{x}_3 \mathbf{x}_4)$ is:

### Related Formula For a geometric progression, the terms can be set as a, ar, ar^2, ar^3$a, ar, ar^2, ar^3$. For three terms A, B, C$A, B, C$ to be in arithmetic progression, they must satisfy: 2B = A + C$2B = A + C$ ### Core Logic Let the elements be x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3$x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3$. After the specified subtractions, the sequence becomes: a - 2, quad ar - 7, quad ar^2 - 9, quad ar^3 - 5$$a - 2, \quad ar - 7, \quad ar^2 - 9, \quad ar^3 - 5$$ Since this sequence is in AP, we form two separate common difference linear linkages: 2(ar - 7) = (a - 2) + (ar^2 - 9) implies 2ar - 14 = ar^2 + a - 11 implies ar^2 - 2ar + a + 3 = 0 quad dots (1)$$2(ar - 7) = (a - 2) + (ar^2 - 9) \implies 2ar - 14 = ar^2 + a - 11 \implies ar^2 - 2ar + a + 3 = 0 \quad \dots (1)$$ 2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) implies 2ar^2 - 18 = ar^3 + ar - 12 implies ar^3 - 2ar^2 + ar + 6 = 0 quad dots (2)$$2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) \implies 2ar^2 - 18 = ar^3 + ar - 12 \implies ar^3 - 2ar^2 + ar + 6 = 0 \quad \dots (2)$$ ### Step 1: Solve the Simultaneous Polynomials Multiply equation (1) by r$r$: ar^3 - 2ar^2 + ar + 3r = 0 quad dots (3)$$ar^3 - 2ar^2 + ar + 3r = 0 \quad \dots (3)$$ Subtract equation (3) from equation (2): (ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0$$(ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0$$ 6 - 3r = 0 implies 3r = 6 implies r = 2$$6 - 3r = 0 \implies 3r = 6 \implies r = 2$$ Substitute r = 2$r = 2$ back into equation (1): a(2)^2 - 2a(2) + a + 3 = 0$$a(2)^2 - 2a(2) + a + 3 = 0$$ 4a - 4a + a + 3 = 0 implies a = -3$$4a - 4a + a + 3 = 0 \implies a = -3$$ ### Step 2: Find the Continuous Product Value The continuous product term is: mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = a cdot ar cdot ar^2 cdot ar^3 = a^4 r^6$$\mathbf{x}_1\mathbf{x}_2\mathbf{x}_3\mathbf{x}_4 = a \cdot ar \cdot ar^2 \cdot ar^3 = a^4 r^6$$ mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = (-3)^4 cdot (2)^6 = 81 times 64 = 5184$$\mathbf{x}_1\mathbf{x}_2\mathbf{x}_3\mathbf{x}_4 = (-3)^4 \cdot (2)^6 = 81 \times 64 = 5184$$ Now divide by 24 as required: frac124(5184) = 216$$\frac{1}{24}(5184) = 216$$ ### Pattern Recognition Notice that multiplying the first AP condition equation by r$r$ perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series