In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the :
(i) first term
(ii) common difference
(iii) sum of the first 20 terms

Answer

To solve the problem step by step, we will use the properties of Arithmetic Progression (A.P.).

### Step 1: Understand the terms of A.P.
The general term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write equations for the given terms
We know:
- The 4th term \( a_4 = 8 \)
- The 6th term \( a_6 = 14 \)

Using the formula for the general term:
1. For the 4th term:
\[ a + 3d = 8 \quad \text{(Equation 1)} \]
2. For the 6th term:
\[ a + 5d = 14 \quad \text{(Equation 2)} \]

### Step 3: Subtract Equation 1 from Equation 2
To eliminate \( a \), we subtract Equation 1 from Equation 2:
\[ (a + 5d) - (a + 3d) = 14 - 8 \]
This simplifies to:
\[ 2d = 6 \]
From this, we can find \( d \):
\[ d = \frac{6}{2} = 3 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now, substitute \( d = 3 \) back into Equation 1:
\[ a + 3(3) = 8 \]
This simplifies to:
\[ a + 9 = 8 \]
Thus, we can solve for \( a \):
\[ a = 8 - 9 = -1 \]

### Step 5: Find the sum of the first 20 terms
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \times (2(-1) + (20-1) \times 3) \]
This simplifies to:
\[ S_{20} = 10 \times (-2 + 19 \times 3) \]
Calculating further:
\[ S_{20} = 10 \times (-2 + 57) = 10 \times 55 = 550 \]

### Final Answers
(i) First term \( a = -1 \)
(ii) Common difference \( d = 3 \)
(iii) Sum of the first 20 terms \( S_{20} = 550 \)