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

Options

A. 32nd term
B. 12th term
C. 22nd term
D. 42nd term

Answer

To find the first negative term of the arithmetic progression (AP) given by 121, 117, 113, ..., we can follow these steps:

### Step 1: Identify the first term (A) and the common difference (D)
- The first term \( A = 121 \)
- The common difference \( D = 117 - 121 = -4 \)

### Step 2: Write the formula for the nth term of an AP
The nth term \( a_n \) of an AP can be expressed as:
\[ a_n = A + (n - 1) \cdot D \]

### Step 3: Set up the inequality for the first negative term
We want to find the first term that is negative, so we set up the inequality:
\[ a_n < 0 \]
Substituting the formula for \( a_n \):
\[ 121 + (n - 1)(-4) < 0 \]

### Step 4: Simplify the inequality
Now, simplify the inequality:
\[ 121 - 4(n - 1) < 0 \]
Distributing the -4:
\[ 121 - 4n + 4 < 0 \]
Combining like terms:
\[ 125 - 4n < 0 \]

### Step 5: Solve for n
Rearranging the inequality:
\[ 125 < 4n \]
Dividing both sides by 4:
\[ n > \frac{125}{4} \]
Calculating \( \frac{125}{4} \):
\[ n > 31.25 \]

### Step 6: Determine the smallest integer n
Since \( n \) must be a positive integer, the smallest integer greater than 31.25 is 32.

### Conclusion
Thus, the first negative term of the AP occurs at the 32nd term.

### Final Answer
The first negative term of the AP is the 32nd term.

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