If P Q R S T is a pentagon, then the resultant of forces P Q ¯ , P T ¯ , Q R ¯ , S R ¯ ,…

Given, PQRST is a pentagon

Let, $\stackrel{-}{p},\stackrel{-}{q},\stackrel{-}{r},\stackrel{-}{s} and \stackrel{-}{t}$ be the position vectors associated with  $\stackrel{-}{P}, \stackrel{-}{Q, }\stackrel{-}{R}, \stackrel{-}{S }and \stackrel{-}{T}$ respectively

Now, $\stackrel{-}{PQ}+\stackrel{-}{PT}+\stackrel{-}{QR}+\stackrel{-}{SR}+\stackrel{-}{TS}+\stackrel{-}{PS}=$

$\Rightarrow (\stackrel{-}{q}-\stackrel{-}{p})+(\stackrel{-}{t}-\stackrel{-}{p})+(\stackrel{-}{r}-\stackrel{-}{q})+(\stackrel{-}{s}-\stackrel{-}{r})+(\stackrel{-}{s}-t)+(\stackrel{-}{s}-\stackrel{-}{p})$

$\Rightarrow 3\stackrel{-}{s}-3\stackrel{-}{p}$

$\Rightarrow 3(\stackrel{-}{s}-\stackrel{-}{p})$

$\Rightarrow 3\stackrel{-}{PS}$.