In an A.P:
given `a_(n) = 4` , d = 2 , `S_(n) = -14 ` , find n and a .

Answer

To solve the problem step by step, we will use the formulas for the nth term and the sum of the first n terms of an arithmetic progression (A.P).

### Given:
- \( a_n = 4 \)
- \( d = 2 \)
- \( S_n = -14 \)

### Step 1: Use the nth term formula
The formula for the nth term of an A.P is given by:
\[ a_n = a + (n - 1) \cdot d \]
Substituting the known values:
\[ 4 = a + (n - 1) \cdot 2 \]

### Step 2: Rearranging the equation
Rearranging the equation gives:
\[ 4 = a + 2n - 2 \]
Adding 2 to both sides:
\[ 6 = a + 2n \]
Thus, we can express \( a \) in terms of \( n \):
\[ a = 6 - 2n \quad \text{(Equation 1)} \]

### Step 3: Use the sum formula
The formula for the sum of the first n terms of an A.P is:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
Substituting the known values:
\[ -14 = \frac{n}{2} \cdot (2a + (n - 1) \cdot 2) \]
This simplifies to:
\[ -14 = \frac{n}{2} \cdot (2a + 2n - 2) \]
Factoring out the 2:
\[ -14 = \frac{n}{2} \cdot 2(a + n - 1) \]
This further simplifies to:
\[ -14 = n(a + n - 1) \]

### Step 4: Substitute \( a \) from Equation 1
Now we substitute \( a \) from Equation 1 into the sum equation:
\[ -14 = n((6 - 2n) + n - 1) \]
Simplifying inside the parentheses:
\[ -14 = n(5 - n) \]
Expanding gives:
\[ -14 = 5n - n^2 \]
Rearranging this into standard quadratic form:
\[ n^2 - 5n - 14 = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the quadratic equation
We will solve the quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -5 \), and \( c = -14 \):
\[ n = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ n = \frac{5 \pm \sqrt{25 + 56}}{2} \]
\[ n = \frac{5 \pm \sqrt{81}}{2} \]
\[ n = \frac{5 \pm 9}{2} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{14}{2} = 7 \)
2. \( n = \frac{-4}{2} = -2 \) (not valid since \( n \) must be positive)

Thus, \( n = 7 \).

### Step 6: Find \( a \)
Now, substitute \( n = 7 \) back into Equation 1 to find \( a \):
\[ a = 6 - 2(7) \]
\[ a = 6 - 14 = -8 \]

### Final Answer:
- \( n = 7 \)
- \( a = -8 \)

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