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Hint: The formula for writing \[{{n}^{th}}\] term of an arithmetic progression is

\[{{n}^{th}}\ term=a+(n-1)d\]

(Where ‘a’ is the first term and‘d’ is the common difference of the arithmetic progression)

We can find that the given series is an arithmetic progression by firstly assuming that the given series is an arithmetic progression. Then if the series follows the rules of an arithmetic progression, then we can conclude that the given is an arithmetic progression.

As mentioned in the question, we have to find whether the series which is given to us is an arithmetic progression or not.

Now, as mentioned in the hint, let us take the given series to be as an arithmetic progression. So, we can say that 1, 4, 7, 10, 13, 16, 19, 22, 25 … is an arithmetic progression.

Now, as given in the hint, if the above series follows the formula for writing \[{{n}^{th}}\] term of an arithmetic progression, then, we can say that it is actually an arithmetic progression.

Now, as we count, we can see that the 9th term of this series is 25, so, we can write as follows

\[{{n}^{th}}\ term=a+(n-1)d\]

(Where ‘a’ is the first term which 1 in this case and ‘d’ is the common difference of the arithmetic progression which is 3 in this case, if we consider it as an arithmetic progression)

\[\begin{align}

& {{9}^{th}}\ term=1+(9-1)3 \\

& 25=1+8\times 3 \\

& 25=1+24 \\

& 25=25 \\

\end{align}\]

As L.H.S and R.H.S are equal, hence, we can say that the given series is actually an arithmetic progression because it follows the rules of an arithmetic progression.

Note: The students can make an error in writing the sum and \[{{n}^{th}}\] term if they might confuse in finding the common difference that is ‘d’ as the value of ‘d’ would be found only on proceeding with the question by taking the common difference as an unknown variable.

Also, another method of doing this question is that we can see in the series that there is a constant difference between two successive terms which is what a property of arithmetic progression is.

\[{{n}^{th}}\ term=a+(n-1)d\]

(Where ‘a’ is the first term and‘d’ is the common difference of the arithmetic progression)

We can find that the given series is an arithmetic progression by firstly assuming that the given series is an arithmetic progression. Then if the series follows the rules of an arithmetic progression, then we can conclude that the given is an arithmetic progression.

__Complete step-by-step answer:__As mentioned in the question, we have to find whether the series which is given to us is an arithmetic progression or not.

Now, as mentioned in the hint, let us take the given series to be as an arithmetic progression. So, we can say that 1, 4, 7, 10, 13, 16, 19, 22, 25 … is an arithmetic progression.

Now, as given in the hint, if the above series follows the formula for writing \[{{n}^{th}}\] term of an arithmetic progression, then, we can say that it is actually an arithmetic progression.

Now, as we count, we can see that the 9th term of this series is 25, so, we can write as follows

\[{{n}^{th}}\ term=a+(n-1)d\]

(Where ‘a’ is the first term which 1 in this case and ‘d’ is the common difference of the arithmetic progression which is 3 in this case, if we consider it as an arithmetic progression)

\[\begin{align}

& {{9}^{th}}\ term=1+(9-1)3 \\

& 25=1+8\times 3 \\

& 25=1+24 \\

& 25=25 \\

\end{align}\]

As L.H.S and R.H.S are equal, hence, we can say that the given series is actually an arithmetic progression because it follows the rules of an arithmetic progression.

Note: The students can make an error in writing the sum and \[{{n}^{th}}\] term if they might confuse in finding the common difference that is ‘d’ as the value of ‘d’ would be found only on proceeding with the question by taking the common difference as an unknown variable.

Also, another method of doing this question is that we can see in the series that there is a constant difference between two successive terms which is what a property of arithmetic progression is.

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