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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