Question

The interior angles of an octagon are in ${\text{A.P.}}$. The smallest angle is $30^\circ$ and the common difference is $30^\circ$. Find the sum of all the angles.

Answer 1

Hint: Here, we will use the formula for sum of A.P. is ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ is the smallest angle, $d$ is the common difference and $n$ is the number of angles. Then, substitute the values of $a$, $d$ and $n$ in expression for ${{\text{S}}_n}$.

Complete step-by-step solution:
Given that an octagon has 8 sides.
We will use the formula for sum of all angles ${{\text{S}}_n}$ is $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ is the smallest angle, $d$ is the common difference and $n$ is the number of angles.

Since it is given that the interior angles of an octagon are in A.P, the angles can be written as $a$, $a + d$, $a + 3d$, $a + 4d$, $a + 5d$, $a + 6d$ and $a + 7d$.

Now we will find the values of $a$, $d$ and $n$.

$a = 30$
$d = 30$
$n = 8$

Substituting these values of $a$, $d$ and $n$ in expression for ${{\text{S}}_n}$, we get

$${{\text{S}}_n} = \dfrac{8}{2}\left[ {2\left( {30} \right) + \left( {8 - 1} \right)30} \right] \\\ = 4\left[ {60 + 7\left( {30} \right)} \right] \\\ = 4\left[ {60 + 210} \right] \\\ = 4\left[ {270} \right] \\\ = 1080 \\\$$

Thus, the sum of all angles of an octagon is $1080^\circ$.

Note: In this question, some students mistakenly write the formula for the sum of all angles. Also, we are supposed to write the values properly to avoid any miscalculation.