Find the A.P. whose second term is 12 and 7th term exceeds the 4th by 15.

Options

A. `7,12,17,22`
B. `8,12,16,20`
C. `9,13,17,21`
D. None

Answer

To find the arithmetic progression (A.P.) whose second term is 12 and the 7th term exceeds the 4th term by 15, we can follow these steps:

### 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 of the A.P. can be expressed as follows:
- First term: \( a \)
- Second term: \( a + d \)
- Fourth term: \( a + 3d \)
- Seventh term: \( a + 6d \)

### Step 2: Use the information given in the problem.
From the problem, we know:
1. The second term is 12:
\[ a + d = 12 \quad \text{(Equation 1)} \]

2. The seventh term exceeds the fourth term by 15:
\[ (a + 6d) - (a + 3d) = 15 \]
Simplifying this gives:
\[ 6d - 3d = 15 \implies 3d = 15 \quad \text{(Equation 2)} \]

### Step 3: Solve Equation 2 for \( d \).
From Equation 2:
\[ 3d = 15 \]
Dividing both sides by 3:
\[ d = 5 \]

### Step 4: Substitute \( d \) back into Equation 1 to find \( a \).
Now substitute \( d = 5 \) into Equation 1:
\[ a + 5 = 12 \]
Subtracting 5 from both sides:
\[ a = 12 - 5 = 7 \]

### Step 5: Write the A.P. using \( a \) and \( d \).
Now that we have \( a = 7 \) and \( d = 5 \), we can write the A.P.:
- First term: \( a = 7 \)
- Second term: \( a + d = 7 + 5 = 12 \)
- Third term: \( a + 2d = 7 + 2 \times 5 = 17 \)
- Fourth term: \( a + 3d = 7 + 3 \times 5 = 22 \)
- Fifth term: \( a + 4d = 7 + 4 \times 5 = 27 \)
- Sixth term: \( a + 5d = 7 + 5 \times 5 = 32 \)
- Seventh term: \( a + 6d = 7 + 6 \times 5 = 37 \)

Thus, the A.P. is:
\[ 7, 12, 17, 22, 27, 32, 37, \ldots \]

### Final Answer:
The arithmetic progression is \( 7, 12, 17, 22, 27, 32, 37, \ldots \)

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