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**Hint:**We solve this problem using the standard representation of four terms of an A.P as

\[\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)\]

Then we use the sum and product of terms of a polynomial of degree 4 that is

If \[p,q,r,s\]are the roots of equation \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0\] then, the sum of roots is

\[\Rightarrow p+q+r+s=\dfrac{-b}{a}\]

The sum of product of roots taken two at a time is

\[\Rightarrow pq+qr+rs+sp+pr+sq=\dfrac{c}{a}\]

The product of roots

\[\Rightarrow pqrs=\dfrac{e}{a}\]

By using the above formulas to given equation we find the value of \['m'\]

**Complete step by step answer:**

We are given that the polynomial equation of degree 4 as

\[\Rightarrow {{x}^{4}}-\left( 3m+2 \right){{x}^{2}}+{{m}^{2}}=0\]

Let us rewrite the above equation by placing the missing terms as

\[\Rightarrow {{x}^{4}}+0{{x}^{3}}-\left( 3m+2 \right){{x}^{2}}+0x+{{m}^{2}}=0\]

We are given that the roots of above equation are in A.P

We know that the standard representation of four terms of an A.P as

\[\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)\]

Let us assume that the roots of given equation as

\[\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)\]

We know that the sum of roots and product of roots of equation of degree 4 that is

If \[p,q,r,s\]are the roots of equation \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0\] then, the sum of roots is

\[\Rightarrow p+q+r+s=\dfrac{-b}{a}\]

The sum of product of roots taken two at a time is

\[\Rightarrow pq+qr+rs+sp+pr+sq=\dfrac{c}{a}\]

The product of roots

\[\Rightarrow pqrs=\dfrac{e}{a}\]

By using the above formulas to given equation we find the value of \['m'\]

Now, by using the sum of roots to given equation we get

\[\begin{align}

& \Rightarrow a-3d+a-d+a+d+a+3d=\dfrac{-0}{1}=0 \\

& \Rightarrow 4a=0 \\

& \Rightarrow a=0 \\

\end{align}\]

Now the roots of given equation will be

\[-3d,-d,d,3d\]

Now by using the sum of product of roots taken two at a time we get

\[\begin{align}

& \Rightarrow \left( -3d\times -d \right)+\left( -d\times d \right)+\left( d\times 3d \right)+\left( 3d\times -3d \right)+\left( -3d\times d \right)+\left( -d\times 3d \right)=\dfrac{-\left( 3m+2 \right)}{1} \\

& \Rightarrow -10{{d}^{2}}=-\left( 3m+2 \right) \\

& \Rightarrow {{d}^{2}}=\dfrac{\left( 3m+2 \right)}{10} \\

\end{align}\]

Now by using the product of roots formula we get

\[\begin{align}

& \Rightarrow -3d\times -d\times d\times 3d=\dfrac{{{m}^{2}}}{1} \\

& \Rightarrow 9{{\left( {{d}^{2}} \right)}^{2}}={{m}^{2}} \\

\end{align}\]

Now, by substituting the value of \['{{d}^{2}}'\] in above equation we get

\[\begin{align}

& \Rightarrow 9{{\left( \dfrac{3m+2}{10} \right)}^{2}}={{m}^{2}} \\

& \Rightarrow 9\left( 9{{m}^{2}}+12m+4 \right)=100{{m}^{2}} \\

& \Rightarrow 19{{m}^{2}}-108m-36=0 \\

\end{align}\]

Now by using the factorisation method that is rewriting the middle in such a way that we get factors then we get

\[\begin{align}

& \Rightarrow 19{{m}^{2}}-114m+6m-36=0 \\

& \Rightarrow 19m\left( m-6 \right)+6\left( m-6 \right)=0 \\

& \Rightarrow \left( m-6 \right)\left( 19m+6 \right)=0 \\

\end{align}\]

We know that if \[a\times b=0\] then either of \[a,b\] will be zero.

By using this result to above equation we get the first term as

\[\Rightarrow m=6\]

Similarly, by taking the second term we get

\[\Rightarrow m=\dfrac{-6}{19}\]

**We are given that \[m>0\] so, the value of \['m'\] is 6.**

**Note:**Students may make mistakes in taking the terms of an A.P.

The general representation of A.P is

\[a,\left( a+d \right),\left( a+2d \right),\left( a+3d \right),.......\]

Students may take these four terms which gives the complex solution.

But if we consider the terms as

\[\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)\]

Here, we get the solution easily because while adding the terms we get only one variable remains which will be easy to solve.

This point needs to be taken care of.

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