The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term of the A.P. is `(1)/(3)`Calculate the first term and the 13th term.

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (A.P.).

### Step 1: Understand the given information
We know that:
1. The sum of the first six terms of the A.P. is 42.
2. The ratio of the 10th term to the 30th term of the A.P. is \( \frac{1}{3} \).

### 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 of an A.P. is given by:
\[ S_n = \frac{n}{2} \cdot [2a + (n-1)d] \]
For \( n = 6 \):
\[ S_6 = \frac{6}{2} \cdot [2a + (6-1)d] = 3 \cdot [2a + 5d] \]
Setting this equal to 42:
\[ 3(2a + 5d) = 42 \]
Dividing both sides by 3:
\[ 2a + 5d = 14 \quad \text{(Equation 1)} \]

### Step 3: Write the expressions for the 10th and 30th terms
The \( n \)-th term of an A.P. is given by:
\[ T_n = a + (n-1)d \]
Thus, for the 10th term:
\[ T_{10} = a + 9d \]
And for the 30th term:
\[ T_{30} = a + 29d \]

### Step 4: Set up the ratio of the 10th term to the 30th term
Given that the ratio of the 10th term to the 30th term is \( \frac{1}{3} \):
\[ \frac{T_{10}}{T_{30}} = \frac{1}{3} \]
Substituting the expressions for \( T_{10} \) and \( T_{30} \):
\[ \frac{a + 9d}{a + 29d} = \frac{1}{3} \]
Cross-multiplying gives:
\[ 3(a + 9d) = 1(a + 29d) \]
Expanding both sides:
\[ 3a + 27d = a + 29d \]
Rearranging terms:
\[ 3a - a = 29d - 27d \]
This simplifies to:
\[ 2a = 2d \]
Thus, we have:
\[ a = d \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Substituting \( a = d \) into Equation 1:
\[ 2a + 5d = 14 \]
Since \( d = a \):
\[ 2a + 5a = 14 \]
This simplifies to:
\[ 7a = 14 \]
Dividing both sides by 7:
\[ a = 2 \]

### Step 6: Find the common difference
Using Equation 2, since \( a = d \):
\[ d = 2 \]

### Step 7: Calculate the 13th term
The 13th term is given by:
\[ T_{13} = a + (13 - 1)d = a + 12d \]
Substituting the values of \( a \) and \( d \):
\[ T_{13} = 2 + 12 \cdot 2 = 2 + 24 = 26 \]

### Final Answers
- The first term \( a = 2 \)
- The 13th term \( T_{13} = 26 \)