The sum of the first six terms of an A.P. is 42. The ratio of the 10th term to the 30th term of A.P. is `1/3` Find the 40th term of the A.P.:

Options

A. `-60`
B. 20
C. 39
D. 80

Answer

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

### Step 1: Understanding the sum of the first six terms of an A.P.
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

Given that the sum of the first six terms \( S_6 = 42 \):
\[ S_6 = \frac{6}{2} \times (2a + (6-1)d) = 42 \]
This simplifies to:
\[ 3(2a + 5d) = 42 \]
Dividing both sides by 3:
\[ 2a + 5d = 14 \quad \text{(Equation 1)} \]

### Step 2: Using the ratio of the 10th term to the 30th term
The \( n \)-th term of an A.P. is given by:
\[ T_n = a + (n-1)d \]
Thus, the 10th term \( T_{10} \) and the 30th term \( T_{30} \) are:
\[ T_{10} = a + 9d \]
\[ T_{30} = a + 29d \]
We are 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:
\[ 3a - a + 27d - 29d = 0 \]
This simplifies to:
\[ 2a - 2d = 0 \quad \Rightarrow \quad a = d \quad \text{(Equation 2)} \]

### Step 3: Substituting \( a = d \) into Equation 1
Now we substitute \( a = d \) into Equation 1:
\[ 2a + 5d = 14 \]
Since \( a = d \):
\[ 2a + 5a = 14 \]
This simplifies to:
\[ 7a = 14 \]
Dividing both sides by 7:
\[ a = 2 \]
Since \( a = d \), we also have:
\[ d = 2 \]

### Step 4: Finding the 40th term of the A.P.
The formula for the \( n \)-th term is:
\[ T_n = a + (n-1)d \]
To find the 40th term \( T_{40} \):
\[ T_{40} = a + 39d \]
Substituting the values of \( a \) and \( d \):
\[ T_{40} = 2 + 39 \times 2 \]
Calculating:
\[ T_{40} = 2 + 78 = 80 \]

### Final Answer
The 40th term of the A.P. is \( \boxed{80} \).