[Hint : Find n for an< 0]
The given A.P. is \(121, 117, 113 \dots\). Here, the first term is \(a = 121\). The common difference is \(d = 117 - 121 = -4\). The formula for the nth term of an A.P. is \(a_n = a + (n-1)d\). Substituting the values, we get \(a_n = 121 + (n-1)(-4)\). Simplifying, \(a_n = 121 - 4n + 4\), which further simplifies to \(a_n = 125 - 4n\). We need to find the first negative term. This means we need to find \(n\) such that \(a_n<0\). So, we set \(a_n<0\): \(125 - 4n<0\) \(125<4n\) \(n>\frac{125}{4}\) \(n>31.25\) Since \(n\) must be an integer representing the term number, the smallest integer greater than 31.25 is 32. Therefore, the 32nd term will be the first negative term of this A.P.