Which term of the AP : 121, 117, 113, ldots i | vedclass

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(B) Given $A.P.$ is $121, 117, 113, \ldots$Here,the first term $a = 121$ and the common difference $d = 117 - 121 = -4$.The $n^{	ext{th}}$ term of an

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$32$

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(B) Given $A.P.$ is $121, 117, 113, \ldots$Here,the first term $a = 121$ and the common difference $d = 117 - 121 = -4$.The $n^{	ext{th}}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.Substituting the values,we get $a_n = 121 + (n - 1)(-4) = 121 - 4n + 4 = 125 - 4n$.We want to find the first negative term,so we set $a_n $125 - 4n $125 $n > \frac{125}{4}$$n > 31.25$Since $n$ must be a natural number,the smallest integer greater than $31.25$ is $32$.Therefore,the $32^{	ext{nd}}$ term is the first negative term of this $A.P.$

Which term of the $AP : 121, 117, 113, \ldots$ is its first negative term?

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