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Hint: Zeros of a polynomial are roots of a polynomial. Since the roots are in A.P., they can be assumed to be (a – d), a and a + d where the common difference between the terms is d.

For a polynomial of degree 3, $a{{x}^{3}}+b{{x}^{3}}+cx+d$_{, }with roots $\alpha ,\beta ,\gamma ,$ the sum of roots is given as $\alpha +\beta +\gamma =\dfrac{-b}{a}$, the sum of roots taken two at a time is given as $\alpha \beta +\beta \gamma +\gamma \delta =\dfrac{c}{a}$ , and the product of the roots is given as $\alpha \beta \gamma =\dfrac{-d}{a}$.Using these concepts we will find the value of k.

__Complete step-by-step answer:__

Now, the polynomial given to us is \[{{x}^{3}}-12{{x}^{2}}+39x+k\] . On comparing $a{{x}^{3}}+b{{x}^{3}}+cx+d$ , we get a = 1, b = -12, c = 39 and d = k. Now, let’s assume the zeros of the polynomial to be (a – d), a and (a + d). So, the sum of roots can be written as

\[\left( a-d \right)+a+\left( a+d \right)=\dfrac{\left( -b \right)}{a}=\dfrac{-\left( -12 \right)}{1}\]

\[\Rightarrow 3a=12\]

\[\Rightarrow a=4\]

Now, the sum of roots taken two at a time is given as \[\left( a+d \right)\left( a \right)+\left( a \right)\left( a-d \right)+\left( a+d \right)\left( a-d \right)=\dfrac{\left( c \right)}{a}=39\]

\[\Rightarrow a\left( a+d+a-d \right)+{{a}^{2}}-{{d}^{2}}=39\]

\[\Rightarrow 3{{a}^{2}}-{{d}^{2}}=39\]

Now, we have a = 4. On substituting, we get $3\times {{\left( 4 \right)}^{2}}-{{d}^{2}}=39$ .

$\Rightarrow 48-{{d}^{2}}=39$

\[\begin{align}

& \Rightarrow {{d}^{2}}=9 \\

& \Rightarrow d=\pm 3 \\

\end{align}\]

In this question, we can take d to be either $3$ or $\ -3$ , as both +d and – d are present symmetrically. Hence, it won’t make a difference.

We take $d=3$.

We now know the value of a and d. We'll use these values to find the roots.

The roots are (a + d), a and a – d.

Therefore, the roots are $4+3=7$, $4$ and $4-3=1$.

We know, the product of roots is given as $\alpha \beta \gamma =\dfrac{-d}{a}$ .

$\Rightarrow \alpha \beta \gamma =-k$

$\Rightarrow \left( a-d \right)\left( a \right)\left( a+d \right)=-k$

$\Rightarrow \left( 1 \right)\left( 4 \right)\left( 7 \right)=-k$

$\Rightarrow k=-28$

Therefore, the required value of k is –28.

Note: In any question, where we have to find series in arithmetic progression, like in this question, we had to first find the roots; we should assume the terms using two variables such that their sum results in one variable.

For example, if we had to find four terms, the terms would be taken as a – 3d, a – d, a + d, a + 3d.

For a polynomial of degree 3, $a{{x}^{3}}+b{{x}^{3}}+cx+d$

Now, the polynomial given to us is \[{{x}^{3}}-12{{x}^{2}}+39x+k\] . On comparing $a{{x}^{3}}+b{{x}^{3}}+cx+d$ , we get a = 1, b = -12, c = 39 and d = k. Now, let’s assume the zeros of the polynomial to be (a – d), a and (a + d). So, the sum of roots can be written as

\[\left( a-d \right)+a+\left( a+d \right)=\dfrac{\left( -b \right)}{a}=\dfrac{-\left( -12 \right)}{1}\]

\[\Rightarrow 3a=12\]

\[\Rightarrow a=4\]

Now, the sum of roots taken two at a time is given as \[\left( a+d \right)\left( a \right)+\left( a \right)\left( a-d \right)+\left( a+d \right)\left( a-d \right)=\dfrac{\left( c \right)}{a}=39\]

\[\Rightarrow a\left( a+d+a-d \right)+{{a}^{2}}-{{d}^{2}}=39\]

\[\Rightarrow 3{{a}^{2}}-{{d}^{2}}=39\]

Now, we have a = 4. On substituting, we get $3\times {{\left( 4 \right)}^{2}}-{{d}^{2}}=39$ .

$\Rightarrow 48-{{d}^{2}}=39$

\[\begin{align}

& \Rightarrow {{d}^{2}}=9 \\

& \Rightarrow d=\pm 3 \\

\end{align}\]

In this question, we can take d to be either $3$ or $\ -3$ , as both +d and – d are present symmetrically. Hence, it won’t make a difference.

We take $d=3$.

We now know the value of a and d. We'll use these values to find the roots.

The roots are (a + d), a and a – d.

Therefore, the roots are $4+3=7$, $4$ and $4-3=1$.

We know, the product of roots is given as $\alpha \beta \gamma =\dfrac{-d}{a}$ .

$\Rightarrow \alpha \beta \gamma =-k$

$\Rightarrow \left( a-d \right)\left( a \right)\left( a+d \right)=-k$

$\Rightarrow \left( 1 \right)\left( 4 \right)\left( 7 \right)=-k$

$\Rightarrow k=-28$

Therefore, the required value of k is –28.

Note: In any question, where we have to find series in arithmetic progression, like in this question, we had to first find the roots; we should assume the terms using two variables such that their sum results in one variable.

For example, if we had to find four terms, the terms would be taken as a – 3d, a – d, a + d, a + 3d.

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