In the formula `t_(n)=(1)/(a+(n-1)d)`, how many auxiliary formulae are possible ad what are they?

Answer

To derive the auxiliary formulas from the given formula \( t_n = \frac{1}{a + (n-1)d} \), we will solve for each variable \( a \), \( d \), and \( n \) in terms of the others.

### Step-by-Step Solution:

1. **Start with the original formula:**
\[ t_n = \frac{1}{a + (n-1)d} \]

2. **Rearranging to find \( a \):**
- Multiply both sides by \( a + (n-1)d \):
\[ t_n(a + (n-1)d) = 1 \]
- Expanding gives:
\[ at_n + (n-1)dt_n = 1 \]
- Isolate \( a \):
\[ at_n = 1 - (n-1)dt_n \]
- Divide by \( t_n \):
\[ a = \frac{1 - (n-1)dt_n}{t_n} \]
- **Auxiliary formula for \( a \)**:
\[ a = \frac{1 - (n-1)dt_n}{t_n} \]

3. **Rearranging to find \( d \):**
- Start again from the original formula:
\[ t_n = \frac{1}{a + (n-1)d} \]
- Multiply both sides by \( a + (n-1)d \):
\[ t_n(a + (n-1)d) = 1 \]
- Rearranging gives:
\[ (n-1)dt_n = 1 - at_n \]
- Isolate \( d \):
\[ d = \frac{1 - at_n}{(n-1)t_n} \]
- **Auxiliary formula for \( d \)**:
\[ d = \frac{1 - at_n}{(n-1)t_n} \]

4. **Rearranging to find \( n \):**
- Start again from the original formula:
\[ t_n = \frac{1}{a + (n-1)d} \]
- Multiply both sides by \( a + (n-1)d \):
\[ t_n(a + (n-1)d) = 1 \]
- Expanding gives:
\[ at_n + (n-1)dt_n = 1 \]
- Rearranging gives:
\[ (n-1)dt_n = 1 - at_n \]
- Isolate \( n \):
\[ n - 1 = \frac{1 - at_n}{dt_n} \]
- Therefore:
\[ n = \frac{1 - at_n}{dt_n} + 1 \]
- **Auxiliary formula for \( n \)**:
\[ n = \frac{1 - at_n}{dt_n} + 1 \]

### Summary of Auxiliary Formulas:
1. \( a = \frac{1 - (n-1)dt_n}{t_n} \)
2. \( d = \frac{1 - at_n}{(n-1)t_n} \)
3. \( n = \frac{1 - at_n}{dt_n} + 1 \)