Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]
Step-by-Step Solution
Key Concept: Use the formula for the nth term of an arithmetic progression, \(a_n = a_1 + (n-1)d\), and set the inequality \(a_n < 0\) to find the smallest integer \(n\) that satisfies it.
1. Identify the first term and common difference:
\[a_1 = 121\]
\[d = 117 - 121 = -4\]
2. Write the general term of the AP:
\[a_n = a_1 + (n-1)d = 121 + (n-1)(-4)\]
Simplify:
\[a_n = 121 - 4(n-1) = 125 - 4n\]
3. Impose the condition for the first negative term:
\[a_n < 0 \Rightarrow 125 - 4n < 0\]
4. Solve the inequality:
\[125 < 4n \Rightarrow n > \frac{125}{4} = 31.25\]
Since \(n\) must be a positive integer, the smallest possible value is \(n = 32\).
5. Verify the 32nd term:
\[a_{32} = 121 + (32-1)(-4) = 121 - 124 = -3\]
The 31st term is still positive (\(a_{31}=1\)), confirming that the 32nd term is indeed the first negative term.
6. Hence, the first negative term occurs at the 32nd position of the AP.
Correct Answer: 32nd term (n = 32)