20 The sum of the first 53 terms of an arithmetic series is 4240 , while the seventh term is equal to 20 . Find the first term and the common difference.

To find the first term and the common difference of the arithmetic series, we can use the formulas related to the sum of an arithmetic series and the nth term. The formula for the sum of the first \( n \) terms of an arithmetic series is: \[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \] Where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. Given: - \( S_{53} = 4240 \) - \( n = 53 \) We can plug in these values: \[ 4240 = \frac{53}{2} \times (2a + 52d) \] Multiplying both sides by 2: \[ 8480 = 53(2a + 52d) \] Now divide both sides by 53: \[ 160 = 2a + 52d \quad (1) \] It is also given that the seventh term is 20: \[ a + 6d = 20 \quad (2) \] Now we have two equations: 1. \( 2a + 52d = 160 \) 2. \( a + 6d = 20 \) From equation (2), we can express \( a \): \[ a = 20 - 6d \quad (3) \] Now substitute equation (3) into equation (1): \[ 2(20 - 6d) + 52d = 160 \] This simplifies to: \[ 40 - 12d + 52d = 160 \] Combining like terms gives: \[ 40 + 40d = 160 \] Subtracting 40 from both sides: \[ 40d = 120 \] Now divide by 40: \[ d = 3 \] Now that we have \( d \), substitute it back into equation (3) to find \( a \): \[ a = 20 - 6(3) = 20 - 18 = 2 \] Thus, the first term \( a \) is 2, and the common difference \( d \) is 3. **Final Answer:** The first term is 2, and the common difference is 3.