$p_n \to r, q_n \to r, \alpha_n , \beta_n \in [0,1]$ and $\alpha_n + \beta_n =1$ for all $n$, prove that $p_n \alpha_n +q_n \beta_n\to r$ please help me solve this question thanks in advance. 

$p_n \to r, q_n \to r, \alpha_n , \beta_n \in [0,1]$ and $\alpha_n + \beta_n =1$ for all $n$, prove that $p_n \alpha_n +q_n \beta_n\to r$.

I am stuck I don't know how should I start solving this question.

You have to use the definition of limit: so, we want that $\forall \varepsilon\  \exists N_0,$ such that $\forall n \geq N_0, |\alpha_n p_n +\beta_n q_n -r| < \varepsilon.$
Now, by definition of limit $\exists N_p, N_q$ such that $\forall n\geq N_p \ |p_n-r|< \varepsilon; \forall n\geq N_q \ |q_n-r|< \varepsilon.$
Now, take $N_0=max\{N_p,N_q\}:$ for every $n\geq N_0$ we have that $|\alpha_n p_n +\beta_n q_n -r|=|\alpha_n (p_n-r) +\beta_n (q_n - r)|\leq |\alpha_n (p_n-r)|+|\beta_n (q_n - r)|\leq \alpha_n \varepsilon+\beta_n \varepsilon=\varepsilon.$ Notice that I've used two times that $\forall n \ \alpha_n+\beta_n=1.$
Thus we have obtained the thesis.
