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Hint: Start with finding out the general term of the A.P. Then put it less than zero to find the first negative term.

According to the question, the given sequence is 121, 117, 113,…..

As we can see, the first term of A.P. is 121 and its common difference is -4. So, we have:

$ \Rightarrow a = 121{\text{ and }}d = - 4$

We know that the general term of A.P. is:

$ \Rightarrow {T_n} = a + \left( {n - 1} \right)d$

Putting the values of $a$ and $d$ from above, we get:

$

\Rightarrow {T_n} = 121 + \left( {n - 1} \right)\left( { - 4} \right), \\

\Rightarrow {T_n} = 121 - 4n + 4, \\

\Rightarrow {T_n} = 125 - 4n \\

$

For the term in the sequence to be negative, we have:

$

\Rightarrow {T_n} < 0, \\

\Rightarrow 125 - 4n < 0, \\

\Rightarrow 4n > 125, \\

\Rightarrow n > \dfrac{{125}}{4}, \\

\Rightarrow n > 31.25 \\

$

Therefore, the first negative term of A.P. is the 32nd term.

Note:

$n = 31$ will be the last positive term of the given A.P. and $n \ne 31.25$. $n$ must be an integer. So, $n = 32$ will be the first negative term.

According to the question, the given sequence is 121, 117, 113,…..

As we can see, the first term of A.P. is 121 and its common difference is -4. So, we have:

$ \Rightarrow a = 121{\text{ and }}d = - 4$

We know that the general term of A.P. is:

$ \Rightarrow {T_n} = a + \left( {n - 1} \right)d$

Putting the values of $a$ and $d$ from above, we get:

$

\Rightarrow {T_n} = 121 + \left( {n - 1} \right)\left( { - 4} \right), \\

\Rightarrow {T_n} = 121 - 4n + 4, \\

\Rightarrow {T_n} = 125 - 4n \\

$

For the term in the sequence to be negative, we have:

$

\Rightarrow {T_n} < 0, \\

\Rightarrow 125 - 4n < 0, \\

\Rightarrow 4n > 125, \\

\Rightarrow n > \dfrac{{125}}{4}, \\

\Rightarrow n > 31.25 \\

$

Therefore, the first negative term of A.P. is the 32nd term.

Note:

$n = 31$ will be the last positive term of the given A.P. and $n \ne 31.25$. $n$ must be an integer. So, $n = 32$ will be the first negative term.