If the zeros of the polynomial …

Given {tex}f(x)=2{ x }^{ 3 }-15{ x }^{ 2 }+37x-30{/tex}

Let the zeroes of the polynomial in A.P  be a-d,a,a+d

Using  the relationship between the zeroes and  coefficients of a cubic polynomial ,we get

Sum of the roots = {tex}a-d+a+a+d=\frac { -\left( -15 \right) }{ 2 } {/tex} 

{tex}\Rightarrow 3a=\frac { 15 }{ 2 } \\ \Rightarrow a=\frac { 5 }{ 2 } {/tex}

Now the product of the roots= {tex}\left( a-d \right) a\left( a+d \right) =\frac { -\left( -30 \right) }{ 2 } {/tex}

{tex}\Rightarrow a\left( { a }^{ 2 }-{ d }^{ 2 } \right) =\frac { 30 }{ 2 } \\ \Rightarrow \frac { 5 }{ 2 } \left( \frac { 25 }{ 4 } -{ d }^{ 2 } \right) =15\\ \Rightarrow \left( \frac { 25 }{ 4 } -{ d }^{ 2 } \right) =15\times \frac { 2 }{ 5 } =6\\ \Rightarrow { d }^{ 2 }=\frac { 25 }{ 4 } -6=\frac { 25-24 }{ 4 } =\frac { 1 }{ 4 } \\ \Rightarrow { d }=\pm \frac { 1 }{ 2 } {/tex}

Hence the roots are {tex}\frac { 5 }{ 2 } -\frac { 1 }{ 2 } ,\frac { 5 }{ 2 } ,\frac { 5 }{ 2 } +\frac { 1 }{ 2 } =2,2.5,3{/tex}