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# Which term of the AP: 121, 117. 113, ….., is its first negative term?

Given:

Given A.P. is $121, 117, 113,…..$

To do:

We have to find which term of the given A.P. is its first negative term.

Solution:

Here,

$a_1=121, a_2=117, a_3=113$

Common difference $d=a_2-a_1=117-121=-4$

The first negative term of the given A.P. $=121-4\times31=121-124=-3$ ($121-4\times30=1$ is the last positive term)

We know that,

nth term $a_n=a+(n-1)d$

Therefore,

$a_{n}=121+(n-1)(-4)$

$-3=121+n(-4)-1(-4)$

$-3-121=-4n+4$

$124+4=4n$

$4n=128$

$n=\frac{128}{4}$

$n=32$

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

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