Chapter 7, Problem 17TP

Solution

To state: The first six terms of the sequence then tell whether the sequence is arithmetic, geometric, or neither if the number of diagonals in a convex polygon is given by the formula $d_n=\frac{1}{2}n\left(n-3\right)$ where n is the number of sides of the polygon $\left(n\ge 3\right)$ .

The resultant six terms are $0,2,5,9,14,20$ and it can be said that the sequence is neither.

Given information:

The number of diagonals in a convex polygon is given by the formula $d_n=\frac{1}{2}n\left(n-3\right)$ where n is the number of sides of the polygon $\left(n\ge 3\right)$ .

Explanation:

Consider the given formula: $d_n=\frac{1}{2}n\left(n-3\right)$

Find the first six terms by substituting the values of n as $3,4,5,6,7,8$ .

  $\begin{array}{c}{d}_3=\frac{1}{2}\cdot 3\left(3-3\right),d_3=0\\ d_4=\frac{1}{2}\cdot 4\left(4-3\right),d_4=\frac{1}{2}\cdot 4,d_4=2\\ ,d_5=\frac{1}{2}\cdot 5\left(5-3\right),d_5=\frac{1}{2}\cdot 5\left(2\right),d_5=5\\ d_6=\frac{1}{2}\cdot 6\left(6-3\right),d_6=\frac{1}{2}\cdot 6\left(3\right),d_6=9\\ d_7=\frac{1}{2}\cdot 7\left(7-3\right),d_7=\frac{1}{2}\cdot 7\left(4\right),d_7=14\\ d_{38}=\frac{1}{2}\cdot 8\left(8-3\right),d_8=\frac{1}{2}\cdot 8\left(5\right),d_8=20\end{array}$

The first six terms are: $0,2,5,9,14,20$

Now check if the sequence is arithmetic or not, for that find the common difference d :

  $\begin{array}{l}2-0=2\\ 5-2=3\\ 9-5=4\end{array}$

It can be seen that the common difference is not same. Therefore, the sequence is not arithmetic.

Now check if the sequence is geometric or not, for that find the common ratio r :

  $\begin{array}{l}r=\frac{2}{0},r=\infty \\ r=\frac{5}{2},r=2.5\end{array}$

It can be seen that the common ratio is not same. Therefore, the sequence is not geometric.

Therefore, it can be said that the sequence is neither.