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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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