Question

The interior angles of a polygon are in arithmetic progression. The smallest angle is \[{{120}^{o}}\] and the common difference is 5. Find the number of sides of the polygon.

Answer 1

Hint:

 First of all consider a ‘n’ sided polygon, then find an A.P with the first term as \[{{120}^{o}}\] and common difference as 5. Now, use the formula for the sum of the n terms of A.P that is \[\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]. Also, the sum of ‘n’ interior angles of the polygon is \[\left( n-2 \right){{180}^{o}}\]. So, equate these values and find the value of n.

Complete step-by-step answer:
Here, we are given that the interior angles of a polygon are in arithmetic progression and the smallest angle is \[{{120}^{o}}\] and the common difference is 5. We have to find the number of sides of the polygon. We know that if the first term of A.P is ‘a’ and common difference is ‘d’, then A.P is given by

\[a,a+d,a+2d,a+3d......\text{ n terms}\]

Similarly, we are given that the interior angles are in A.P and the first term, and the common differences are \[{{120}^{o}}\] and 5 respectively. So, we get the A.P as,

\[{{120}^{o}},{{120}^{o}}+{{5}^{o}},{{120}^{o}}+\left( 2\times 5 \right),{{120}^{o}}+\left( 3\times 5 \right)......\text{ n terms}\]

\[{{120}^{o}},{{125}^{o}},{{130}^{o}},{{135}^{o}}......\text{ n terms}\]

Above is the sequence of interior angles of a polygon. Let us take the total sides of the polygon to be ‘n’. So, these would be ‘n’ interior angles as well. We know that the sum of n terms of an A.P is given by

\[\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right).....\left( i \right)\]

where ‘n’ is the number of terms, ‘a’ is the first term and ‘d’ is the common difference.

By substituting \[a={{120}^{o}}\], d = 5 in equation (i), we get,

Sum of interior angles of n sided polygon

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

\[=\dfrac{n}{2}\left( 5n+235 \right)....\left( ii \right)\]

We also know that by geometry, the sum of the interior angles of n sided polygon is given by
\[\left( n-2 \right){{180}^{o}}.....\left( iii \right)\]

So by equating RHS of the equation (ii) and (iii), we get,

\[\Rightarrow \dfrac{n}{2}\left( 5n+{{235}^{o}} \right)=\left( n-2 \right){{180}^{o}}\]

By simplifying the above equation, we get,

\[\dfrac{\left( 5{{n}^{2}}+235n \right)}{2}=180n-360\]

By multiplying 2 on both sides of the above equation, we get,

\[\Rightarrow 5{{n}^{2}}+235n=360n-720\]

By transposing all the terms to LHS of the above equation, we get,

\[5{{n}^{2}}+235n-360n+720=0\]

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

By dividing the whole equation by 5, we get,

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

We can write the above equation as,

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

\[n\left( n-16 \right)-9\left( n-16 \right)=0\]

\[\left( n-16 \right)\left( n-9 \right)=0\]

So, we get,

n – 16 = 0; n – 9 =0

n = 16; n = 9

So, here we get the number of sides of the polygon as 16 or 9.

Note: In this question, students must note that if it would be given that the given polygon is a convex, then we won’t take n = 16 because, for n = 16, some angles would be greater than \[{{180}^{o}}\] and we know that in a convex polygon, the interior angles are always less than \[{{180}^{o}}\]. Also, students must remember that in a sided polygon, the sum of the interior angles is given by \[\left( n-2 \right){{180}^{o}}\].