Question

4 Calculate: \( \frac{4+6+8 \ldots+806}{5+10+15+\ldots+625} \)

Ask by Mcdonald Alexander. in South Africa
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The ratio is \(\frac{3,\!618}{875}\).

Solution

To calculate the ratio \(\frac{4 + 6 + 8 + \ldots + 806}{5 + 10 + 15 + \ldots + 625}\), we'll evaluate both the numerator and the denominator separately and then simplify the resulting fraction. ### **Numerator: \(4 + 6 + 8 + \ldots + 806\)** This is an arithmetic sequence where: - **First term (\(a_1\))** = 4 - **Common difference (\(d\))** = 2 - **Last term (\(a_n\))** = 806 **Number of terms (\(n\))**: \[ a_n = a_1 + (n-1)d \\ 806 = 4 + (n-1) \times 2 \\ 806 = 4 + 2n - 2 \\ 806 = 2n + 2 \\ 804 = 2n \\ n = 402 \] **Sum of the numerator (\(S_n\))**: \[ S_n = \frac{n}{2} \times (a_1 + a_n) \\ S_n = \frac{402}{2} \times (4 + 806) \\ S_n = 201 \times 810 \\ S_n = 162{,}810 \] ### **Denominator: \(5 + 10 + 15 + \ldots + 625\)** This is also an arithmetic sequence where: - **First term (\(a_1\))** = 5 - **Common difference (\(d\))** = 5 - **Last term (\(a_m\))** = 625 **Number of terms (\(m\))**: \[ a_m = a_1 + (m-1)d \\ 625 = 5 + (m-1) \times 5 \\ 625 = 5m \\ m = 125 \] **Sum of the denominator (\(S_m\))**: \[ S_m = \frac{m}{2} \times (a_1 + a_m) \\ S_m = \frac{125}{2} \times (5 + 625) \\ S_m = \frac{125}{2} \times 630 \\ S_m = 125 \times 315 \\ S_m = 39{,}375 \] ### **Ratio of the Sums** \[ \frac{S_n}{S_m} = \frac{162{,}810}{39{,}375} \] **Simplifying the Fraction:** \[ \frac{162{,}810 \div 15}{39{,}375 \div 15} = \frac{10{,}854}{2{,}625} \\ \frac{10{,}854 \div 3}{2{,}625 \div 3} = \frac{3{,}618}{875} \] The fraction \(\frac{3{,}618}{875}\) cannot be simplified further as 3,618 and 875 share no common factors besides 1. ### **Final Answer** \[ \frac{3,\!618}{875} \] Which is the simplified ratio of the two sums. **Decimal Representation (Optional):** \[ \frac{3{,}618}{875} \approx 4.1343 \] --- **Answer:** The ratio is \(\dfrac{3,\!618}{875}\).

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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.

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