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Which term of the AP: 121,117,113,..., is its first negative term. Find \[n\] for \[{a_n} =0\]

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Hint: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. Use the formula of the general term to find the value of n in this question.
Using the formula of general term in this question, \[{T_n} = a + \left( {n - 1} \right)d = 0\]
We get,
\[{T_n} = 121 + \left( {n - 1} \right)\left( { - 4} \right) = 0\]
Therefore, on solving this becomes,
\[n = \dfrac{{121}}{4} + 1\]
On further solving,
We get,
\[n = 31.25\]

Note: Since 31.25th term does not exist, make sure to round off the number to the nearest number.
This clearly means the value of can be taken roughly as \[32\],
Therefore,
\[n \approx 32\]