Sequences and series: Give the difference between consecutive angles and smallest angle. Find the number of sides Here is the question:
The difference between any two consecutive interior angles of a polygon is 5 degrees. If the smallest angle  is 120 degrees. Find the number of sides the polygon has.
I think the answer is 9. But I thought it will be an even number since only then the figure can be made(I am not sure about this since I got 9)
Can you help me with this question?
 A: It seems to me there is no solution. Proof:
The sum of the exterior angles of any polygon is $360^\circ$, so we need to find a set of numbers $\theta_0,\theta_1,\dots,\theta_k=\theta_0$ such that $|\theta_i-\theta_{i-1}|=5^\circ$ and $\sum_1^k\theta_i=360^\circ$. We also know that $\min\{180^\circ-\theta_i\}=120^\circ$, so $\theta_i\leq 60^\circ$ and $(\exists i)[\theta_i=60^\circ]$. WLOG, we can let $\theta_0=60^\circ$. Noting that $\theta_i/5^\circ=n_i\in\mathbb Z$, the constraint $|n_i-n_{i-1}|=1$ means that if $n_i$ is even, $n_{i+1}$ is odd and vice-versa so that $n_i$ is even iff $i$ is even since $n_0=12$ by induction.
Thus, if $k$ is odd, then $n_k=n_0$ is odd and even, a contradiction. Thus $k$ is even. Additionally, since $n_0+n_1,n_2+n_3,\dots$ are all odd, $\sum_1^k n_i$ will be odd if $k/2$ is odd, which is a contradiction since $\sum_1^k n_i=72$. Thus $k$ is a multiple of $4$.
Now $$n_i\leq 12\Rightarrow\sum_1^k n_i=72\leq12k\Rightarrow k\geq 6,$$ and $|n_i-n_{i-1}|=1\Rightarrow n_i\geq12-\min(i,n-i)\Rightarrow$
$$72=\sum_1^k n_i=n_0+n_{k/2}+\sum_1^{k/2-1} (n_i+n_{n-i})\geq 24-k/2+2\sum_1^{k/2-1} (12-i)=12k-(k/2)^2,$$
so $k\leq12(2-\sqrt2)\approx7.02$. There are no multiples of 4 in the range $k\in[6,7.02]$, so this has no solution.
A: The sum of all angles in a polygon is given by the formula....$180(n-2)$....also we know that angles are in A.P( according to question) so again using the formula for that $\frac{n}{2}(2a + (n-1)d )$.....equating these we get....  $$180(n-2) = n/2( 120 + 120 + 5(n-1))$$
If you simplify, you get
\begin{align}
5n^2 -125n +720 &=0\\
n^2 -25n +144 &=0\\
(n -16)(n-9) &=0
\end{align}
Thereofore, $n =16$ or $n =9$
