In an A.P. the third term is four times the first term, and the sixth term is 17 , find the series.

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 \( n \)-th term of an A.P. is given by the formula:
\[ T_n = a + (n - 1)d \]

### Step 2: Write the equations based on the given information.
From the problem, we know:
1. The third term \( T_3 \) is four times the first term:
\[ T_3 = a + 2d = 4a \]
2. The sixth term \( T_6 \) is 17:
\[ T_6 = a + 5d = 17 \]

### Step 3: Set up the equations.
From the first equation:
\[ a + 2d = 4a \]
Rearranging gives:
\[ 3a = 2d \quad \text{(Equation 1)} \]

From the second equation:
\[ a + 5d = 17 \quad \text{(Equation 2)} \]

### Step 4: Solve for \( d \) in terms of \( a \).
From Equation 1, we can express \( d \) in terms of \( a \):
\[ d = \frac{3a}{2} \]

### Step 5: Substitute \( d \) into Equation 2.
Now substitute \( d \) into Equation 2:
\[ a + 5\left(\frac{3a}{2}\right) = 17 \]
This simplifies to:
\[ a + \frac{15a}{2} = 17 \]
Combining the terms gives:
\[ \frac{2a + 15a}{2} = 17 \]
\[ \frac{17a}{2} = 17 \]

### Step 6: Solve for \( a \).
Multiplying both sides by 2:
\[ 17a = 34 \]
Dividing by 17:
\[ a = 2 \]

### Step 7: Find \( d \) using \( a \).
Now substitute \( a = 2 \) back into Equation 1 to find \( d \):
\[ 3(2) = 2d \implies 6 = 2d \implies d = 3 \]

### Step 8: Write the series.
Now that we have \( a = 2 \) and \( d = 3 \), we can write the series:
\[ \text{The series is } 2, 5, 8, 11, 14, 17, \ldots \]

### Final Answer:
The series is \( 2, 5, 8, 11, 14, 17, \ldots \)

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