Write a recursive formula for the following explicit formula, and find the first term of the sequence. $$ \begin{array}{l}a_{n}=46+41 \mathrm{n} \ a_{1}=\square \ a_{n}=\square \ \text { (Simplify your answers.) }\end{array} $$

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To convert the explicit formula $$ a_n = 46 + 41n $$ into a recursive formula and find the first term of the sequence, follow these steps:

### 1. Find the First Term ($$ a_1 $$):
Substitute $$ n = 1 $$ into the explicit formula:
$$
a_1 = 46 + 41(1) = 46 + 41 = 87
$$

### 2. Develop the Recursive Formula:
The explicit formula $$ a_n = 46 + 41n $$ represents an **arithmetic sequence** with a common difference $$ d $$.

- **Common Difference ($$ d $$)**: 
  $$
  d = 41
  $$

- **Recursive Formula Structure**:
  For an arithmetic sequence, the recursive formula is:
  $$
  a_n = a_{n-1} + d
  $$
  
- **Applying the Common Difference**:
  $$
  a_n = a_{n-1} + 41 \quad \text{for } n \geq 2
  $$

### 3. Summary of the Recursive Formula:
$$
\begin{cases}
a_1 = 87 \
a_n = a_{n-1} + 41 & \text{for } n \geq 2
\end{cases}
$$

### **Final Answer:**
$$
\begin{aligned}
a_1 &=\, 87 \
a_n &=\, a_{n-1} + 41 \quad \text{for } n \geq 2
\end{aligned}
$$
First term: 87 Recursive formula: $$ a_n = a_{n-1} + 41 $$ for $$ n \geq 2 $$

Write a recursive formula for the following explicit formula, and find the first term of the sequence. \[ \begin{array}{l}a_{n}=46+41 \mathrm{n} \\ a_{1}=\square \\ a_{n}=\square \\ \text { (Simplify your answers.) }\end{array} \]

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