For an arithmetic sequence, $a_{40}=56$. If the common difference is 6 , find: \[ a_{1}= \] the sum of the first 58 terms = Question Help: Message instructor Submit Question

Step 1 ：We are given that the 40th term of an arithmetic sequence is 56 and the common difference is 6. We are asked to find the first term of the sequence and the sum of the first 58 terms.

Step 2 ：We can find the first term of the sequence using the formula for the nth term of an arithmetic sequence: \(a_n = a_1 + (n-1)d\). In this case, \(a_{40} = 56\), \(d = 6\), and \(n = 40\). Solving for \(a_1\), we get \(a_1 = a_{40} - (n-1)d = 56 - (40-1)*6 = -178\).

Step 3 ：We can find the 58th term of the sequence using the same formula: \(a_{58} = a_1 + (58-1)*6 = -178 + (58-1)*6 = 164\).

Step 4 ：We can find the sum of the first 58 terms using the formula for the sum of an arithmetic sequence: \(S_n = \frac{n}{2}(a_1 + a_n)\). In this case, \(S_{58} = \frac{58}{2}(-178 + 164) = -406\).

Step 5 ：Final Answer: The first term of the sequence is \(a_1 = \boxed{-178}\) and the sum of the first 58 terms is \(S_{58} = \boxed{-406}\).