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:

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

[Where $a_n$ is the $n^{th}$ term, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference.]

Step 3: Set up the inequality to find the first negative term.

We are looking for the first term that is less than zero. Therefore, we set $a_n < 0$:

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

Substitute the known values $a = 121$ and $d = -4$ into the inequality:

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

Step 4: Solve the inequality for $n$.

$121 - 4n + 4 < 0$ [Distributing $-4$ into the parentheses]

$125 - 4n < 0$ [Combining like terms $121 + 4 = 125$]

$-4n < -125$ [Subtracting $125$ from both sides]

$4n > 125$ [Multiplying by $-1$ reverses the inequality sign]

$n > \frac{125}{4}$ [Dividing both sides by $4$]

$n > 31.25$

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

Since $n$ must be a positive integer representing the position of a term in the sequence, and we require the smallest integer $n$ such that $n > 31.25$, we conclude that $n = 32$.

Step 6: Verification (Optional but recommended).

Calculate the $31^{st}$ term: $a_{31} = 121 + (31 - 1)(-4) = 121 + 30(-4) = 121 - 120 = 1$. (This is positive)

Calculate the $32^{nd}$ term: $a_{32} = 121 + (32 - 1)(-4) = 121 + 31(-4) = 121 - 124 = -3$. (This is the first negative term)

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