Write a recursive formula for the following explicit formula, and find the first term of \( t \) sequence. \( a_{n}=31+41 \mathrm{n} \) \( a_{1}=\square \) \( a_{n}=\square \) (Simplify your answers.)

To convert the explicit formula \( a_n = 31 + 41n \) into a recursive formula and find the first term \( a_1 \), follow these steps: 1. **Find the First Term (\( a_1 \)):** Substitute \( n = 1 \) into the explicit formula: \[ a_1 = 31 + 41 \times 1 = 31 + 41 = 72 \] 2. **Derive the Recursive Formula:** A recursive formula defines each term based on the previous term. To find the relationship between \( a_n \) and \( a_{n-1} \): \[ a_n = 31 + 41n \] \[ a_{n-1} = 31 + 41(n-1) = 31 + 41n - 41 = 41n - 10 \] Subtract the two equations to find the difference: \[ a_n - a_{n-1} = (31 + 41n) - (41n - 10) = 31 + 10 = 41 \] This shows that each term increases by 41 from the previous term. Therefore, the recursive formula is: \[ a_n = a_{n-1} + 41 \quad \text{for } n \geq 2 \] **Final Answer:** \[ a_1 = 72 \] \[ a_n = a_{n-1} + 41 \]