Convex polygon sides

We know that:

• $\color{#3D99F6}{a_1 = 100}, \color{#D61F06}{d = 4^\circ}$

• The sum of $n$ terms of an A.P: $S_n = \dfrac{n}{2}(a_1 + a_n)$

• The $n$ th term of an AP: $a_n=a_1+(n-1)d$

• $S_n = 180^\circ(n-2)$ as the $\text{RHS}$ is the formula for the sum of interior angles in a polygon.

From substitution:

$\implies S_n=\dfrac{n}{2}(\color{#3D99F6}{100}+a_n)$

$\implies S_n=\dfrac{n}{2}(100+\color{#3D99F6}{100}+(n-1)(\color{#D61F06}{4}))$

$\implies S_n=\dfrac{n}{2}(196+4n) = 2n^2 + 98n$

Now we can use the fourth fact to get the equation: $2n^2 + 98n = 180n - 360 \implies 2n^2 - 82n + 360 = 0$ .

Solving for this quadratic yields $n = 5, 36$ but upon checking, only $\boxed{5}$ works.

we know that the sum of interior angles of a convex polygon is (n-2) $180^{\circ}$ . we are given that a=100 , d=4

$\frac{n}{2}$ {2(100)+(n-1)4}=(n-2)180

$n^{2}$ -41n+180=0 n=5,36 but interior angle can not be more than 180 so n=36 is rejected therefore n=5