If the interior angles of a polygon are in A.P. with common difference `5^(@)` and the smallest angle `120^(@)` , then the number of sides of the polygon is:

Options

A. 9 or 16
B. 9
C. 13
D. 3 or 16

Answer

To find the number of sides of the polygon given that the interior angles are in arithmetic progression (A.P.) with a common difference of \(5^\circ\) and the smallest angle being \(120^\circ\), we can follow these steps:

### Step 1: Define the angles in A.P.
Let the smallest angle \(A = 120^\circ\) and the common difference \(D = 5^\circ\). The angles of the polygon can be expressed as:
- First angle: \(A = 120^\circ\)
- Second angle: \(A + D = 120^\circ + 5^\circ = 125^\circ\)
- Third angle: \(A + 2D = 120^\circ + 2 \times 5^\circ = 130^\circ\)
- And so on...

The \(n^{th}\) angle can be expressed as:
\[ \text{Angle}_n = A + (n-1)D = 120^\circ + (n-1) \times 5^\circ \]

### Step 2: Find the sum of the interior angles of the polygon
The sum of the interior angles \(S\) of a polygon with \(n\) sides is given by the formula:
\[ S = (n - 2) \times 180^\circ \]

### Step 3: Calculate the sum of angles in A.P.
The sum of the angles in A.P. can also be calculated using the formula for the sum of an arithmetic series:
\[ S = \frac{n}{2} \times (\text{First term} + \text{Last term}) \]
The last term (the \(n^{th}\) angle) can be calculated as:
\[ \text{Last term} = 120^\circ + (n-1) \times 5^\circ = 120^\circ + 5n - 5 = 115^\circ + 5n \]
Thus, the sum \(S\) becomes:
\[ S = \frac{n}{2} \times (120 + (115 + 5n)) = \frac{n}{2} \times (235 + 5n) \]

### Step 4: Set the two expressions for the sum equal to each other
Equating the two expressions for \(S\):
\[ (n - 2) \times 180 = \frac{n}{2} \times (235 + 5n) \]

### Step 5: Simplify and solve for \(n\)
Expanding both sides:
\[ 180n - 360 = \frac{n(235 + 5n)}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 360n - 720 = n(235 + 5n) \]
Rearranging gives:
\[ 360n - 720 = 235n + 5n^2 \]
\[ 5n^2 - 125n + 720 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 5\), \(b = -125\), and \(c = 720\):
\[ n = \frac{125 \pm \sqrt{(-125)^2 - 4 \cdot 5 \cdot 720}}{2 \cdot 5} \]
\[ n = \frac{125 \pm \sqrt{15625 - 14400}}{10} \]
\[ n = \frac{125 \pm \sqrt{225}}{10} \]
\[ n = \frac{125 \pm 15}{10} \]

Calculating the two possible values:
1. \(n = \frac{140}{10} = 14\)
2. \(n = \frac{110}{10} = 11\)

### Step 7: Verify the values
Since the angles must be less than \(180^\circ\), we check:
- For \(n = 14\): The largest angle would be \(120 + (14-1) \times 5 = 120 + 65 = 185^\circ\) (not valid).
- For \(n = 11\): The largest angle would be \(120 + (11-1) \times 5 = 120 + 50 = 170^\circ\) (valid).

Thus, the number of sides of the polygon is \(n = 11\).

### Final Answer
The number of sides of the polygon is **11**.