If a, b, c are in arithmetic progression and x, y, z are in geometric progression, and a/x = b/y = c/z = k, then which of the following is true?
Correct: A
Given a, b, c are in AP, so 2b = a + c. Given x, y, z are in GP, so y^2 = xz. Also, a/x = b/y = c/z = k. So a = kx, b = ky, c = kz. Since a, b, c are in AP: 2b = a + c → 2ky = kx + kz → 2y = x + z (since k ≠ 0). This means x, y, z are also in arithmetic progression! So x, y, z are in AP. Check: 2y = x + z, which is the condition for AP. Therefore, Option A is correct: x, y, z are in AP. Note that x, y, z were given to be in GP, and we now find they are also in AP. This implies x, y, z are equal (since only a constant sequence is both AP and GP with non-zero terms). From GP: y^2 = xz. From AP: 2y = x+z. For positive terms, this implies x = y = z. So the common ratio is 1.