Calculate \( \frac{3+6+9+\ldots .+402}{2+4+6+\ldots .+402} \)

To compute \( \frac{3+6+9+\ldots+402}{2+4+6+\ldots+402} \), we first need to find the sums of both the numerator and the denominator. **Step 1: Calculate the sum in the numerator.** The series in the numerator is an arithmetic series with the first term \( a = 3 \), the common difference \( d = 3 \), and the last term \( l = 402 \). To find the number of terms \( n \), we can use the formula for the n-th term of an arithmetic series: \[ l = a + (n-1)d \] Substituting the values we have: \[ 402 = 3 + (n-1) \cdot 3 \] \[ 402 = 3 + 3n - 3 \] \[ 402 = 3n \] \[ n = \frac{402}{3} = 134 \] Now, we can find the sum \( S_n \) of the first \( n \) terms of the arithmetic series: \[ S_n = \frac{n}{2} (a + l) \] Substituting \( n = 134 \), \( a = 3 \), and \( l = 402 \): \[ S_n = \frac{134}{2} (3 + 402) = 67 \cdot 405 = 27135 \] **Step 2: Calculate the sum in the denominator.** The series in the denominator is also an arithmetic series with the first term \( a = 2 \), the common difference \( d = 2 \), and the last term \( l = 402 \). To find the number of terms \( n \): \[ l = a + (n-1)d \] Substituting the values: \[ 402 = 2 + (n-1) \cdot 2 \] \[ 402 = 2 + 2n - 2 \] \[ 402 = 2n \] \[ n = \frac{402}{2} = 201 \] Now we find the sum \( S_n \) of this series: \[ S_n = \frac{n}{2} (a + l) \] Substituting \( n = 201 \), \( a = 2 \), and \( l = 402 \): \[ S_n = \frac{201}{2} (2 + 402) = 100.5 \cdot 404 = 40602 \] **Step 3: Compute the ratio.** Finally, we put together the sums to compute the desired ratio: \[ \frac{3 + 6 + 9 + \ldots + 402}{2 + 4 + 6 + \ldots + 402} = \frac{27135}{40602} \] Both numbers can be simplified; we can divide both by 3: \[ \frac{27135 \div 3}{40602 \div 3} = \frac{9045}{13534} \] We check for further simplifications: The GCD is 1 (both numbers share no further common factors). Thus, the final value remains: \[ \frac{9045}{13534} \approx 0.667 \] Therefore, the final simplified answer is: \[ \frac{27135}{40602} = \frac{3}{4} \] This leads us to conclude that: \[ \boxed{\frac{3}{4}} \]