Let's solve this step-by-step.

Problem:

The sequence is given as $2, 17, 42, 77, 122$, and the $n$-th term is expressed as: $T_n = pn^2 + q,$ where $p$ and $q$ are integers. We are tasked to determine the value of: $\frac{p-q}{p+q}.$

Step 1: Formulate equations using the given sequence

We know the first five terms correspond to $n = 1, 2, 3, 4, 5$. Substituting $n = 1$, $n = 2$, and $n = 3$, we can create equations for $p$ and $q$:

1. For $n = 1$:
   $T_1 = p(1)^2 + q = 2 \quad \Rightarrow \quad p + q = 2.$

2. For $n = 2$:
   $T_2 = p(2)^2 + q = 17 \quad \Rightarrow \quad 4p + q = 17.$

3. For $n = 3$:
   $T_3 = p(3)^2 + q = 42 \quad \Rightarrow \quad 9p + q = 42.$

Step 2: Solve for $p$ and $q$

From the first equation: $q = 2 - p.$

Substitute $q = 2 - p$ into the second equation: $4p + (2 - p) = 17 \quad \Rightarrow \quad 3p + 2 = 17 \quad \Rightarrow \quad 3p = 15 \quad \Rightarrow \quad p = 5.$

Using $p = 5$ in $q = 2 - p$: $q = 2 - 5 = -3.$

Thus, $p = 5$ and $q = -3$.

Step 3: Compute $\frac{p-q}{p+q}$

Substitute $p = 5$ and $q = -3$ into $\frac{p-q}{p+q}$: $\frac{p-q}{p+q} = \frac{5 - (-3)}{5 + (-3)} = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4.$

Final Answer:

$\boxed{b) \, 4}$

Would you like a detailed verification or have other questions? 😊

Related Questions:

1. Can you explain why the quadratic formula for $T_n$ fits this sequence?
2. What would happen if $T_n$ was linear instead of quadratic?
3. How can the value of $\frac{p+q}{p-q}$ be computed?
4. Are there any sequences with similar $p$ and $q$ values?
5. How does changing $q$ affect the sequence?

Tip:

Always verify results by plugging values back into the original equation to ensure consistency.