Which term of the AP: 121, 117, 113, …, is its first negative term?
Hint: Find \(n\) for \(a_n < 0\).
32nd term
Step 1: Identify the first term and common difference.
The AP is: \(121, 117, 113, \ldots\)
First term: \(a = 121\)
Common difference: \(d = 117 - 121 = -4\)
Step 2: Write the nth term.
For an AP, the nth term is:
\[a_n = a + (n - 1)d\]
So here:
\[a_n = 121 + (n - 1)(-4)\]
\[a_n = 121 - 4(n - 1)\]
\[a_n = 121 - 4n + 4 = 125 - 4n\]
Step 3: Use the condition that the term is negative.
We want the first term that is negative, so we need:
\[a_n < 0\]
Substitute \(a_n = 125 - 4n\):
\[125 - 4n < 0\]
Step 4: Solve the inequality.
\[125 < 4n\]
\[n > \dfrac{125}{4} = 31.25\]
Since \(n\) must be a positive integer, the smallest integer greater than 31.25 is:
\[n = 32\]
Step 5: Conclusion.
The 32nd term is the first negative term of the AP.