If the 8th term of an A.P. is 31 and the 15th term is 16 more than the 11th term, find the A.P.

Options

A. 15, 19, 23, 27......
B. 3 , 7 , 11 , 15 .......
C. 7 , 11 , 15,19 .......
D. 11 , 15, 19, 23 .......

Answer

To find the Arithmetic Progression (A.P.) based on the given conditions, we can follow these steps:

### Step 1: Understand the formula for the nth term of an A.P.
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n - 1) \cdot d \]
where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up equations based on the given information.
From the question, we know:
1. The 8th term is 31:
\[ T_8 = a + 7d = 31 \]
This can be labeled as **Equation 1**.

2. The 15th term is 16 more than the 11th term:
\[ T_{15} = a + 14d \]
\[ T_{11} = a + 10d \]
Therefore, we can write:
\[ a + 14d = (a + 10d) + 16 \]
Simplifying this gives:
\[ a + 14d = a + 10d + 16 \]
\[ 14d - 10d = 16 \]
\[ 4d = 16 \]
This can be labeled as **Equation 2**.

### Step 3: Solve for the common difference \( d \).
From Equation 2:
\[ 4d = 16 \]
Dividing both sides by 4:
\[ d = 4 \]

### Step 4: Substitute \( d \) back into Equation 1 to find \( a \).
Using Equation 1:
\[ a + 7d = 31 \]
Substituting \( d = 4 \):
\[ a + 7 \cdot 4 = 31 \]
\[ a + 28 = 31 \]
Subtracting 28 from both sides:
\[ a = 31 - 28 \]
\[ a = 3 \]

### Step 5: Write the A.P. using \( a \) and \( d \).
Now that we have \( a \) and \( d \):
- First term \( a = 3 \)
- Common difference \( d = 4 \)

The A.P. can be expressed as:
\[ a, a + d, a + 2d, a + 3d, \ldots \]
This gives us:
- First term: \( 3 \)
- Second term: \( 3 + 4 = 7 \)
- Third term: \( 3 + 2 \cdot 4 = 11 \)
- Fourth term: \( 3 + 3 \cdot 4 = 15 \)

Thus, the A.P. is:
\[ 3, 7, 11, 15, \ldots \]

### Final Answer:
The Arithmetic Progression is \( 3, 7, 11, 15, \ldots \)

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