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

To solve the expression \( \frac{3+6+9+\ldots+402}{2+4+6+\ldots+402} \), we can evaluate the numerator and denominator separately. **Numerator:** The series \( 3 + 6 + 9 + \ldots + 402 \) is an arithmetic series where the first term \( a = 3 \), the common difference \( d = 3 \), and the last term \( l = 402 \). We can find the number of terms \( n \) in the series using the formula for the \( n \)-th term: \[ l = a + (n-1)d \] Substituting the known values: \[ 402 = 3 + (n-1)3 \] Simplifying this: \[ 402 = 3 + 3n - 3 \] \[ 402 = 3n \] \[ n = \frac{402}{3} = 134 \] Next, we can calculate the sum \( S_n \) of the first \( n \) terms of an arithmetic series using the formula: \[ S_n = \frac{n}{2} (a + l) \] Substituting the values we found: \[ S_{134} = \frac{134}{2} (3 + 402) = 67 \times 405 = 27135 \] **Denominator:** The series \( 2 + 4 + 6 + \ldots + 402 \) is also an arithmetic series where the first term \( a = 2 \), the common difference \( d = 2 \), and the last term \( l = 402 \). Finding the number of terms \( m \): \[ 402 = 2 + (m-1)2 \] Simplifying: \[ 402 = 2 + 2m - 2 \] \[ 402 = 2m \] \[ m = \frac{402}{2} = 201 \] Calculating the sum \( S_m \): \[ S_{201} = \frac{201}{2} (2 + 402) = \frac{201}{2}(404) = 201 \times 202 = 40502 \] **Putting it all together:** Now, we can compute the original expression: \[ \frac{3 + 6 + 9 + \ldots + 402}{2 + 4 + 6 + \ldots + 402} = \frac{27135}{40502} \] To simplify \( \frac{27135}{40502} \), we can divide both the numerator and denominator by their greatest common divisor (gcd). After calculating: The gcd of 27135 and 40502 is 67. Thus: \[ \frac{27135 \div 67}{40502 \div 67} = \frac{405}{605} = \frac{81}{121} \] Final answer: \[ \frac{81}{121} \]