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Is 0 a term of the AP: \( 31,28,25, \ldots \) ? Justify your answer.
Given:
Given AP is \( 31,28,25, \ldots \)
To do:
We have to check whether 0 is a term of the given AP.
Solution:
Let $0$ be the $n$th term of the given AP.
$a_n = 0$
In the given sequence,
$a_1=31, a_2=28, a_3=25$
$d=a_2-a_1=28-31=-3$
If $0$ is a term of the given AP then $n$ should be a positive integer.
Therefore,
$n$th term of the AP $a_n=a+(n-1)d$
$0=31+(n-1)(-3)$
$3(n-1)=31$
$3n-3=31$
$3n=31+3$
$n=\frac{34}{3}$
Here, $n$ is not a positive integer.
Therefore, $0$ is not a term of the given AP.
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