In an A.P:
given a = 8 , `a_(n)= 62, S_(n) = 210` ,find n and d .

Answer

To solve the problem step by step, we will use the formulas for the nth term and the sum of the first n terms of an arithmetic progression (A.P.).

### Given:
- First term \( a = 8 \)
- nth term \( a_n = 62 \)
- Sum of the first n terms \( S_n = 210 \)

### Step 1: Use the formula for the nth term of an A.P.
The formula for the nth term of an A.P. is given by:
\[ a_n = a + (n - 1) \cdot d \]
Substituting the known values:
\[ 62 = 8 + (n - 1) \cdot d \]
This simplifies to:
\[ 62 - 8 = (n - 1) \cdot d \]
\[ 54 = (n - 1) \cdot d \quad \text{(Equation 1)} \]

### Step 2: Use the formula for the sum of the first n terms of an A.P.
The formula for the sum of the first n terms is:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
Substituting the known values:
\[ 210 = \frac{n}{2} \cdot (2 \cdot 8 + (n - 1) \cdot d) \]
This simplifies to:
\[ 210 = \frac{n}{2} \cdot (16 + (n - 1) \cdot d) \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 420 = n \cdot (16 + (n - 1) \cdot d) \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2
From Equation 1, we have:
\[ d = \frac{54}{n - 1} \]
Substituting this into Equation 2:
\[ 420 = n \cdot \left(16 + (n - 1) \cdot \frac{54}{n - 1}\right) \]
This simplifies to:
\[ 420 = n \cdot (16 + 54) \]
\[ 420 = n \cdot 70 \]
Now, solving for \( n \):
\[ n = \frac{420}{70} = 6 \]

### Step 4: Find the value of \( d \)
Now that we have \( n = 6 \), we can substitute it back into Equation 1 to find \( d \):
\[ 54 = (6 - 1) \cdot d \]
\[ 54 = 5d \]
\[ d = \frac{54}{5} = 10.8 \]

### Final Answer:
- \( n = 6 \)
- \( d = 10.8 \)

---