Question 1

We need to find the term number where the value of the AP becomes negative for the first time.

Step 1 — Find the first term and common difference

Let's identify the first term ($a$) and the common difference ($d$) of the given AP. The first term is $a = \mathbf{121}$. We find the common difference by subtracting a term from its succeeding term.

$d = 117 - 121$

$\boxed{-4}$

Step 2 — Set up and solve the inequality

We want to find the first term $a_n$ that is less than zero. Let's write the formula for the $n$-th term of an AP. The formula is $a_n = a + (n-1)d$. We need to find $n$ such that $a_n < 0$.

$a + (n-1)d < 0$

Let's substitute the values of $a$ and $d$.

$121 + (n-1)(-4) < 0$

Now, we will simplify the inequality.

$121 - 4n + 4 < 0$

$125 - 4n < 0$

We need to isolate $n$.

$125 < 4n$

Let's divide both sides by $\mathbf{4}$.

$\frac{125}{4} < n$

$31.25 < n$

This means $n$ must be greater than $\mathbf{31.25}$. Since $n$ must be a whole number (a term number), the smallest integer value for $n$ is $\mathbf{32}$.

$\boxed{n = 32}$

Answer

The 32nd term of the AP is its first negative term.