A sequence $t_{1}, t_{2}, \ldots, t_{n}, \ldots$ is defined as follows:

$$
\begin{aligned}
& t_{1}=14 \\
& t_{k}=24-5 t_{k-1}, \text { for each } k \geq 2
\end{aligned}
$$

For every positive integer $n, t_{n}$ can be expressed as $t_{n}=p \cdot q^{n}+r$, where $p, q$ and $r$ are constants. The value of $p+q+r$ is
(A) -5
(B) -3
(C) 3
(D) 17
(E) 31
A sequence $t_{1}, t_{2}, \ldots, t_{n}, \ldots$ is defined as follows:

$$
\begin{aligned}
& t_{1}=14 \\
& t_{k}=24-5 t_{k-1}, \text { for each } k \geq 2
\end{aligned}
$$

For every positive integer $n$, $t_{n}$ can be expressed as $t_{n}=p \cdot q^{n}+r$, where $p, q$ and $r$ are constants. The value of $p+q+r$ is
(A) -5
(B) -3
(C) 3
(D) 17
(E) 31

## Solution 1

Since $t_{n}=p \cdot q^{n}+r$ for every $n \geq 1$, then

$$
\begin{aligned}
& t_{1}=p q+r \\
& t_{2}=p q^{2}+r \\
& t_{3}=p q^{3}+r
\end{aligned}
$$

However, $t_{1}=14, t_{2}=24-5\left(t_{1}\right)=24-5(14)=-46$, and $t_{3}=24-5 t_{2}=24-5(-46)=254$.

So $p q+r=14$

$$
p q^{2}+r=-46
$$

$$
p q^{3}+r=254
$$

Subtracting (2) - (1) yields $p q^{2}-p q=p q(q-1)=-60$

Subtracting (3) - (2) yields $p q^{3}-p q^{2}=p q^{2}(q-1)=300$

Dividing (5) by (4) gives $q=-5$.

Substituting back into (1) and (2)

$$
\begin{aligned}
-5 p+r & =14 \\
25 p+r & =-46
\end{aligned}
$$

Adding $5 \times$ (1) to (2) yields $6 r=24$ so $r=4$ and thus $p=-2$.

Therefore $t_{n}=-2(-5)^{n}+4$.

So $p+q+r=-2-5+4=-3$.

## Solution 2

Substituting $t_{n}=p \cdot q^{n}+r$ and $t_{n-1}=p \cdot q^{n-1}+r$ into the difference equation $t_{n}=24-5 t_{n-1}$, we get

$$
\begin{aligned}
p \cdot q^{n}+r & =24-5\left(p \cdot q^{n-1}+r\right) \\
p \cdot q^{n}+5 p q^{n-1} & =24-5 r-r \\
p \cdot q^{n-1}(q+5) & =24-6 r
\end{aligned}
$$

Now the right side is independent of $n$, so the left side must be as well, thus $q+5=0$ or $q=-5$. (Clearly, $p \neq 0$ or else $t_{n}$ is constant).

So $24-6 r=0 \Rightarrow r=4$.

Therefore $t_{n}=p(-5)^{n}+4$, so $t_{1}=14=-5 p+4 \Rightarrow p=-2$.

Therefore $p+q+r=-3$.

ANSWER: (B)