Question:

Step 1: Let the three roots of the equation $x^3 - 12x^2 + 39x - 28 = 0$ in arithmetic progression be $a-d$, $a$, and $a+d$, where $a$ is the middle term and $d$ is the common difference. Step 2: Apply Vieta's formulas for the sum of the roots. The sum of the roots is given by $-( \text{coefficient of } x^2 ) / ( \text{coefficient of } x^3 )$. $$ (a-d) + a + (a+d) = -(-12)/1 $$ $$ 3a = 12 $$ $$ a = 4 $$ Step 3: Apply Vieta's formulas for the sum of the products of the roots taken two at a time. The sum of the products of the roots taken two at a time is given by $( \text{coefficient of } x ) / ( \text{coefficient of } x^3 )$. $$ (a-d)a + a(a+d) + (a-d)(a+d) = 39/1 $$ $$ a^2 - ad + a^2 + ad + a^2 - d^2 = 39 $$ $$ 3a^2 - d^2 = 39 $$ Substitute the value $a=4$ into this equation: $$ 3(4^2) - d^2 = 39 $$ $$ 3(16) - d^2 = 39 $$ $$ 48 - d^2 = 39 $$ $$ d^2 = 48 - 39 $$ $$ d^2 = 9 $$ $$ d = \pm 3 $$ Step 4: Verify the result using Vieta's formulas for the product of the roots. The product of the roots is given by $-( \text{constant term} ) / ( \text{coefficient of } x^3 )$. $$ (a-d)a(a+d) = -(-28)/1 $$ $$ a(a^2 - d^2) = 28 $$ Substitute $a=4$ and $d^2=9$: $$ 4(4^2 - 9) = 28 $$ $$ 4(16 - 9) = 28 $$ $$ 4(7) = 28 $$ $$ 28 = 28 $$ The values are consistent. The common difference is $\pm 3$.