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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,

\[n \approx 32\]

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,

\[n \approx 32\]

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