If $\sum_{i=1}^na_i /n \to 1$, then $\sum_{i=1}^n a_iX_i /n \to 1$ holds where $X_n$ iid RVs, $EX_1=1$, and $(a_n)$ is a bdd real seq.

Question

Let $X_n$ be iid RVs, $EX_1=1$, and $(a_n)_{n\geq1}$ be a bounded real sequence.

Show if $\sum_{i=1}^na_i /n \to_{n \to \infty}1$, then $$\sum_{i=1}^n a_iX_i /n \to_{n \to \infty} 1 \ \text{a.s.}$$ holds.

What I have tried:

Fix $\epsilon >0$.

ETS $\ \lim _n P(\sup_{k \geq n} |\sum_{i=1}^k a_iX_i /k-1|> \epsilon)=0$.

Now, $$ P(\sup_{k \geq n} |\sum_{i=1}^k a_iX_i /k-1|> \epsilon) \leq \\ P(\sup_{k \geq n} |\sum_{i=1}^k a_iX_i /k-\sum_{i=1}^ka_i /k|> \epsilon/2)+P(\sup_{k \geq n} |\sum_{i=1}^ka_i /k-1|> \epsilon/2)$$where the second therm is zero whenever n is large.

I want to show 1st term=$P(\sup_{k \geq n} |\sum_{i=1}^k a_i(X_i-1) /k|> \epsilon/2) \to_{n \to \infty} 0$ ,but I cannot see further. (Maybe some SLLN?)

Any hint is appreciated.