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

Given the AP series is 121, 117, 113, . . .,
Thus, first term, a = 121
Common difference, d = 117-121= -4

By the \(n^{th}\) term formula,
\(a_n = a+(n -1)d\)
Therefore,
\(a_n = 121+(n-1)(-4)\)
= 121-4n+4
=125-4n

To find the first negative term of the series, \(a_n < 0\)
Therefore,

\(\Rightarrow \) 125-4n < 0
\(\Rightarrow \) 125 < 4n
\(\Rightarrow \) n>125/4
\(\Rightarrow \) n>31.25

Therefore, the first negative term of the series is \(32^{nd}\) term.