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We are given the difference between two consecutive angles of a polygon as 5Â°.

The smallest angle of the given polygon = 120Â°

So, according to the question, the second smallest angle = 120Â° + 5Â° = 125Â°

Similarly, the third angle will be 125Â° + 5Â°= 130Â°

Therefore, the angles form an A. P. as 120Â°, 125Â°, 130Â°, â€¦

In this A. P., the first term: a = 120Â°

The common difference: d = 5Â°

So, we can use the sum of n terms of an A. P. given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

$

\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \\

\Rightarrow {S_n} = \dfrac{n}{2}\left[ {\left( {2 \times {{120}^ \circ }} \right) + \left( {n - 1} \right){5^ \circ }} \right] \\

$

Now, we know that the sum of all angles of a polygon is given by: SP = ${180^ \circ }(n - 2)$

Both of the equations must be equal because the sum of all the angles of a polygon is definite irrespective of the method used.

$

\Rightarrow {S_n} = {S_p} \\

\Rightarrow \dfrac{n}{2}\left[ {2\left( {120} \right) + \left( {n - 1} \right)5} \right] = 180\left( {n - 2} \right) \\

$

$

\Rightarrow n\left[ {240 + 5n - 5} \right] = 360\left( {n - 2} \right) \\

\Rightarrow 240n + 5{n^2} - 5n = 360n - 720 \\

\Rightarrow 5{n^2} - 125n + 720 = 0 \\

$

Now, we will solve this quadratic equation for the values of n.

$

\Rightarrow 5\left( {{n^2} - 25 + 144} \right) = 0 \\

\Rightarrow {n^2} - 25n + 144 = 0 \\

\Rightarrow {n^2} - 16n - 9n + 144 = 0 \\

\Rightarrow n\left( {n - 16} \right) - 9\left( {n - 16} \right) = 0 \\

\Rightarrow \left( {n - 9} \right)\left( {n - 16} \right) = 0 \\

\Rightarrow n = 9or16 \\

$

We get two values of n and two different numbers of sides in a polygon canâ€™t be acceptable.

So, let us check for the value of n.

For n = 16, $a_{16}$ = a + 15d = 120 + 75 = 195 which is not possible as 195 > 180.