Q1:

Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find $n$ for $a_n < 0$]

Given: An Arithmetic Progression (AP) with terms $121, 117, 113, \dots$

To Find: The value of $n$ such that the $n^{th}$ term ($a_n$) is the first negative term of the sequence.

Step 1: Identify the parameters of the Arithmetic Progression.

The general form of an AP is $a, a+d, a+2d, \dots$, where $a$ is the first term and $d$ is the common difference.

From the given sequence:

First term ($a$) = $121$

Common difference ($d$) = $a_2 - a_1 = 117 - 121 = -4$

Step 2: State the formula for the $n^{th}$ term of an AP.

The formula for the $n^{th}$ term of an AP is given by:

$a_n = a + (n - 1)d$

Substituting the known values:

$a_n = 121 + (n - 1)(-4)$

$a_n = 121 - 4n + 4$

$a_n = 125 - 4n$

Step 3: Formulate the inequality to find the first negative term.

We are looking for the first negative term, which implies we need to solve for $n$ where $a_n < 0$:

$125 - 4n < 0$

Step 4: Solve the inequality for $n$.

Subtract $125$ from both sides:

$-4n < -125$

Divide both sides by $-4$. [Note: When dividing or multiplying an inequality by a negative number, the inequality sign reverses]:

$n > \frac{-125}{-4}$

$n > 31.25$

Step 5: Determine the integer value of $n$.

Since $n$ must be a positive integer representing the position of a term in the sequence, and $n > 31.25$, the smallest integer value for $n$ is $32$.

Step 6: Verification (Optional but recommended).

Calculate the $31^{st}$ term and the $32^{nd}$ term to ensure the result is correct:

For $n = 31$: $a_{31} = 125 - 4(31) = 125 - 124 = 1$ (This is the last positive term)

For $n = 32$: $a_{32} = 125 - 4(32) = 125 - 128 = -3$ (This is the first negative term)

Final Answer: The $32^{nd}$ term of the AP is its first negative term.