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**Hint**: We know that the series is an A.P. when the difference of the consecutive terms are common and difference is checked by taking any term from the series and subtract this term from the successive term of the series. Then check for every consecutive terms given in the option and if all the pairs of consecutive terms of the series are producing a common difference then the series is in A.P. This is how we are going to check the options and then arrive at the conclusion which option is not an A.P.

**:**

__Complete step-by-step answer__We know that A.P. means arithmetic progression. And the series which is an A.P. having common difference meaning if you take any two consecutive terms of an A.P. then you are always getting the same difference. So, we are checking the options given in the above problem by finding the difference of the consecutive terms. The way in which we are calculating the common difference is by taking any term and subtracting the successive term from this term.

Checking option (a) we get,

(a) 13, 8, 3, -2, -7, -12

Subtracting 13 from 8 we get.

$ 8-13=-5 $

Subtracting 8 from 3 we get,

$ 3-8=-5 $

Subtracting 3 from -2 we get,

$ \begin{align}

& 3-\left( -2 \right) \\

& =3+2=5 \\

\end{align} $

Similarly on checking the other options, we have found the difference of -5 as common.

Hence, option (a) is an A.P. with a common difference of -5.

Checking option (b) in the above question we get,

(b) 10.8, 11.2, 11.6, 12, 12.4

Subtracting 10.8 from 11.2 we get,

$ 11.2-10.8=0.4 $

Subtracting 11.2 from 11.6 we get,

$ 11.6-11.2=0.4 $

Similarly, you can check the other terms also and will get the common difference as 0.4.

Hence, this option series is an A.P. with a common difference of 0.4.

Checking option (c) we get,

(c) $ 8\dfrac{1}{7},18\dfrac{2}{7},28\dfrac{3}{7},48\dfrac{4}{7},58\dfrac{5}{7} $

First of all we are converting improper fraction into proper fraction as follows:

$ \begin{align}

& 8\dfrac{1}{7}=8+\dfrac{1}{7} \\

& \Rightarrow \dfrac{56+1}{7}=\dfrac{57}{7} \\

\end{align} $

Similarly, we can convert all the improper fractions given in the question to proper fraction.

$ \dfrac{57}{7},\dfrac{128}{7},\dfrac{199}{7},\dfrac{340}{7},\dfrac{411}{7} $

Subtracting $ \dfrac{57}{7} $ from $ \dfrac{128}{7} $ we get,

$ \begin{align}

& \dfrac{128}{7}-\dfrac{57}{7} \\

& =\dfrac{128-57}{7}=\dfrac{71}{7} \\

\end{align} $

Subtracting $ \dfrac{128}{7} $ from $ \dfrac{199}{7} $ we get,

$ \begin{align}

& \dfrac{199}{7}-\dfrac{128}{7} \\

& =\dfrac{199-128}{7}=\dfrac{71}{7} \\

\end{align} $

Similarly, you can check other terms of this series and will find that difference is common in the series and is $ \dfrac{71}{7} $ .

Hence, this option series is an A.P.

Checking option (d) we get,

(d) $ 8\dfrac{3}{23},11\dfrac{7}{23},14\dfrac{9}{23},17\dfrac{12}{23} $

First of all we are converting improper fraction into proper fraction as follows:

$ \begin{align}

& 8\dfrac{3}{23}=8+\dfrac{3}{23} \\

& \Rightarrow \dfrac{184+3}{23}=\dfrac{187}{23} \\

\end{align} $

Similarly, we can convert all the improper fractions given in the question to proper fractions.

$ \dfrac{187}{23},\dfrac{260}{23},\dfrac{331}{23},\dfrac{403}{23} $

Subtracting $ \dfrac{187}{23} $ from $ \dfrac{260}{23} $ we get,

$ \begin{align}

& \dfrac{260}{23}-\dfrac{187}{23} \\

& =\dfrac{260-187}{23}=\dfrac{73}{23} \\

\end{align} $

Subtracting $ \dfrac{260}{23} $ from $ \dfrac{331}{23} $ we get,

$ \begin{align}

& \dfrac{331}{23}-\dfrac{260}{23} \\

& =\dfrac{331-260}{23}=\dfrac{71}{23} \\

\end{align} $

As you can see that difference is not common so this option series is not an A.P.

**So, the correct answer is “Option D”.**

**Note**: There is a point to be noted in solving this problem or any multiple choice question, if you have to check the options and then mark the correct answer then if in the question it is given that only single option is correct and you are getting second option as correct then don’t check the other options to kill your time. For e.g., in this problem, our first three options are not the correct option so in exam don’t solve the last option mark the last option as correct because it’s a single correct answer question and as we have checked the first three options are incorrect so the last option will be correct.

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