Sequences and Series 2 Question 19
20.
The interior angles of a polygon are in arithmetic progression. The smallest angle is $120^{\circ}$ and the common difference is $5^{\circ}$. Find the number of sides of the polygon.
(1980, 3M)
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Answer:
Correct Answer: 20. (9)
Solution:
- Since, angles of polygon are in an AP.
$\therefore$ Sum of all angles
$ =(n-2) \times 180^{\circ}=\frac{n}{2}{2\left(120^{\circ}\right)+(n-1) 5^{\circ} } $
$ \begin{aligned} & \Rightarrow \quad 5 n^{2}-125 n+720=0 \\ & \Rightarrow \quad n^{2}-25 n+144=0 \\ & \Rightarrow \quad(n-9)(n-16)=0 \\ & \Rightarrow \quad n=9,16 \end{aligned} $
If $n=9$, then largest angle $=a+8 d=160^{\circ}$
Again, if $n=16$, the $n$ largest angle
$ =a+15 d=120^{\circ}+75=195^{\circ} $
which is not possible.
[since, any angle of polygon cannot be $>180^{\circ}$ ]
Hence, $n=9$
[neglecting $n=16$ ]
