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

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

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

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

Answer

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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.