4 Calculate: \( \frac{4+6+8 \ldots+806}{5+10+15+\ldots+625} \)
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To solve the expression \( \frac{4+6+8+\ldots+806}{5+10+15+\ldots+625} \), we need to first calculate the sums in the numerator and the denominator. **Numerator**: The series \( 4 + 6 + 8 + \ldots + 806 \) is an arithmetic series where the first term \( a = 4 \), the last term \( l = 806 \), and the common difference \( d = 2 \). To find the number of terms \( n \), we use the formula for the \( n \)-th term of an arithmetic series: \[ l = a + (n-1)d \] Plugging in our values: \[ 806 = 4 + (n-1) \cdot 2 \] \[ 806 - 4 = (n-1) \cdot 2 \implies 802 = (n-1) \cdot 2 \implies n - 1 = 401 \implies n = 402 \] The sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by: \[ S_n = \frac{n}{2} (a + l) \] Thus, \[ S = \frac{402}{2} (4 + 806) = 201 \cdot 810 = 162810 \] **Denominator**: The series \( 5 + 10 + 15 + \ldots + 625 \) is also an arithmetic series where the first term \( a = 5 \), the last term \( l = 625 \), and the common difference \( d = 5 \). To find the number of terms \( n \), we use the same formula: \[ 625 = 5 + (n-1) \cdot 5 \] \[ 625 - 5 = (n-1) \cdot 5 \implies 620 = (n-1) \cdot 5 \implies n - 1 = 124 \implies n = 125 \] The sum \( S_n \): \[ S_n = \frac{n}{2} (a + l) = \frac{125}{2} (5 + 625) = \frac{125}{2} \cdot 630 = 125 \cdot 315 = 39375 \] Now, we can calculate the overall expression: \[ \frac{162810}{39375} \] Calculating the division gives: \[ \frac{162810 \div 39375}{39375 \div 39375} = \frac{4.1384 \ldots }{1} \approx 4.137 \] However, simplifying it exactly, we can also do the division: \[ \frac{162810}{39375} = 4.1379 \] To finalize, rounding down gives us: \[ \boxed{4} \] as our final answer.