24. Let $\left\{x_{n}\right\}_{n=1}^{\infty}$ be a sequence of real numbers such that $x_{1}=3, x_{2}=24$ and
$$
x_{n+2}=\frac{1}{4} x_{n+1}+\frac{3}{4} x_{n}
$$
for every positive integers $n$. Determine the value of $\lim _{n \rightarrow \infty} x_{n}$.
24. Answer: 15
For each positive integer $n$, we have
$$
x_{n+2}-x_{n+1}=-\frac{3}{4}\left(x_{n+1}-x_{n}\right)
$$
and so we have $x_{n+1}-x_{n}=\left(-\frac{3}{4}\right)^{n-1}\left(x_{2}-x_{1}\right)=21\left(-\frac{3}{4}\right)^{n-1}$.
Therefore, $x_{n}=x_{1}+21 \sum_{k=1}^{n-1}\left(-\frac{3}{4}\right)^{k-1}$ for all positive integers $n \geq 3, \cdots$ (1)
Now, we let $n \rightarrow \infty$ in (1) to conclude that $\lim _{n \rightarrow \infty} x_{n}=x_{1}+21 \sum_{k=1}^{\infty}\left(-\frac{3}{4}\right)^{k-1}=3+\frac{21}{1+0.75}=15$.