MATH SOLVE

5 months ago

Q:
# The measures of the angles of a convex polygon form an arithmetic sequence. The least measurement in the sequence is 128º. The greatest measurement is 172º. Find the number of sides in this polygon given that the sum of the measures is equal to 180(n - 2) degrees. a. 8 c. 22 b. 16 d. 12

Accepted Solution

A:

[tex]\bf \qquad \qquad \textit{sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n}{2}(a_1+a_n)\quad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\ a_n=n^{th}~value\\
----------\\
\stackrel{least}{a_1}=128\\\\
\stackrel{greatest}{a_n}=172
\end{cases}[/tex]

[tex]\bf S_n=\cfrac{n}{2}(128~+~172)\implies S_n=\cfrac{n}{2}(300)\implies S_n=150n \\\\\\ \textit{but we also know that sum of all interior angles is }180(n-2) \\\\\\ \stackrel{\textit{sum of the angles}}{S_n}=\stackrel{\textit{sum of the angles}}{180(n-2)}=150n \\\\\\ 180n-360=150n\implies 30n=360\implies n=\cfrac{360}{30}\implies n=12[/tex]

[tex]\bf S_n=\cfrac{n}{2}(128~+~172)\implies S_n=\cfrac{n}{2}(300)\implies S_n=150n \\\\\\ \textit{but we also know that sum of all interior angles is }180(n-2) \\\\\\ \stackrel{\textit{sum of the angles}}{S_n}=\stackrel{\textit{sum of the angles}}{180(n-2)}=150n \\\\\\ 180n-360=150n\implies 30n=360\implies n=\cfrac{360}{30}\implies n=12[/tex]