The sum of the 2nd term and the 7th term of an A.P. is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P.

Answer

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

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \).

The terms can be expressed as:
- 2nd term \( A_2 = a + d \)
- 7th term \( A_7 = a + 6d \)
- 15th term \( A_{15} = a + 14d \)
- 8th term \( A_8 = a + 7d \)

### Step 2: Set up the first equation
According to the problem, the sum of the 2nd term and the 7th term is 30:
\[ A_2 + A_7 = 30 \]
Substituting the expressions for \( A_2 \) and \( A_7 \):
\[ (a + d) + (a + 6d) = 30 \]
This simplifies to:
\[ 2a + 7d = 30 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation
The problem also states that the 15th term is one less than twice the 8th term:
\[ A_{15} = 2A_8 - 1 \]
Substituting the expressions for \( A_{15} \) and \( A_8 \):
\[ a + 14d = 2(a + 7d) - 1 \]
Expanding the right side:
\[ a + 14d = 2a + 14d - 1 \]
Now, we can simplify this by subtracting \( 14d \) from both sides:
\[ a = 2a - 1 \]
Rearranging gives:
\[ -a = -1 \quad \Rightarrow \quad a = 1 \quad \text{(Equation 2)} \]

### Step 4: Substitute \( a \) back into Equation 1
Now that we have \( a = 1 \), we can substitute this value into Equation 1:
\[ 2(1) + 7d = 30 \]
This simplifies to:
\[ 2 + 7d = 30 \]
Subtracting 2 from both sides:
\[ 7d = 28 \]
Dividing by 7 gives:
\[ d = 4 \quad \text{(Equation 3)} \]

### Step 5: Find the terms of the A.P.
Now that we have both \( a \) and \( d \):
- First term \( a = 1 \)
- Common difference \( d = 4 \)

We can now find the first few terms of the A.P.:
- 1st term: \( A_1 = a = 1 \)
- 2nd term: \( A_2 = a + d = 1 + 4 = 5 \)
- 3rd term: \( A_3 = A_2 + d = 5 + 4 = 9 \)
- 4th term: \( A_4 = A_3 + d = 9 + 4 = 13 \)

Thus, the A.P. can be expressed as:
\[ 1, 5, 9, 13, \ldots \]

### Final Answer
The required Arithmetic Progression (A.P.) is:
\[ \text{A.P.} = 1, 5, 9, 13, \ldots \]