Let the first term of the arithmetic progression (AP) be \( a = 3 \), and let the common difference be \( d \). The sum of the first \( n \) terms of an AP is given by: \[ S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]. \] It is given that the sum of the first 8 terms is twice the sum of the first 5 terms: \[ S_8 = 2S_5. \] Substituting the formula for \( S_n \): \[ \frac{8}{2} \left[ 2a + (8 - 1)d \right] = 2 \cdot \frac{5}{2} \left[ 2a + (5 - 1)d \right]. \] Simplifying both sides: \[ 4 \left[ 2a + 7d \right] = 5 \left[ 2a + 4d \right]. \] Expanding: \[ 8a + 28d = 10a + 20d. \] Simplifying: \[ 8a - 10a + 28d - 20d = 0, \] \[ -2a + 8d = 0. \] Substituting \( a = 3 \): \[ -2(3) + 8d = 0, \] \[ -6 + 8d = 0. \] Solving for \( d \): \[ 8d = 6, \] \[ d = \frac{3}{4}. \] The common difference is \( \frac{3}{4} \).