The interior angles of a polygon are in A.P. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the number of sides of the polygon be $n$.The sum of the interior angles of a polygon with $n$ sides is $(n - 2) 	imes 180^o$.Since the angle

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the number of sides of the polygon be $n$.The sum of the interior angles of a polygon with $n$ sides is $(n - 2) 	imes 180^o$.Since the angles are in $A.P.$ with first term $a = 120^o$ and common difference $d = 5^o$,the sum of the angles is given by $\frac{n}{2}[2a + (n - 1)d]$.Equating the two expressions:$\frac{n}{2}[2(120) + (n - 1)5] = (n - 2)180$$n[240 + 5n - 5] = 360(n - 2)$$5n^2 + 235n = 360n - 720$$5n^2 - 125n + 720 = 0$Dividing by $5$:$n^2 - 25n + 144 = 0$$(n - 9)(n - 16) = 0$So,$n = 9$ or $n = 16$.If $n = 16$,the largest angle is $T_{16} = a + 15d = 120^o + 15(5^o) = 120^o + 75^o = 195^o$.Since an interior angle of a convex polygon must be less than $180^o$,$n = 16$ is rejected.Therefore,the number of sides is $n = 9$.

The interior angles of a polygon are in $A.P.$ If the smallest angle is $120^o$ and the common difference is $5^o$,then the number of sides is

Similar Questions

What is the sum of all two-digit numbers which give a remainder of $4$ when divided by $6$?

The number of terms common to the two $A$.$P$.'s $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ is

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an arithmetic progression are $1/a$,$1/b$,and $1/c$ respectively,then $ab(p - q) + bc(q - r) + ca(r - p) = \dots$

If $a, b, c, d, e$ are in an arithmetic progression,then what is the value of $a - 4b + 6c - 4d + e$?

If the first term of an $A.P.$ is $3$ and the sum of its first four terms is equal to one-fifth of the sum of the next four terms,then the sum of the first $20$ terms is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module