Question

The interior angles of a polygon are in AP. If the smallest angle is ${120}^{\circ }$ & the common difference is $5^{\circ }$, then the number of sides in the polygon are

• A. $7$
• B. $9$
• C. $16$
• D. none of these

Answer 1

The correct option is B $9$

Sum of interior angle of a polygon of $n$ sides$=(2n-4)\pi /2=(n-2)\pi =(n-2){180}^{\circ }$

Given $a={120}^{\circ },d=5^{\circ }$

$\therefore S_n=(n-2){180}^o$

$⟹\frac{n}{2}[2\times 120+(n-1\left)5\right]=(n-2)180$

$⟹n^2-25n+144=0$

$⟹n^2-16n-9n+144=0$

$⟹n(n-16)-9(n-16)=0$

$⟹(n-16)(n-9)=0$

Then $n=16or9$

When $n=16,$ then $T_n=a+(16-1)d$

$=120+15\times 5$

$=120+75={195}^{\circ }$

that is not possible, as the interior angle can not be greater than ${180}^{\circ }$

$\therefore$ $n=9$

Hence, the number of sides in this polygon is $9$.