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# If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in A.P., then its common difference is:A) $\pm 1$B) $\pm 2$C) $\pm 3$D) $\pm 4$  Verified
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Hint:We use the concept of the highest power of any equation is the number of the roots of the equation. Since in our problem we have the highest power three. Hence the given equation has three roots. These are in A.P. we solve the problem by using the properties of A.P.
Let us consider, $\alpha ,\beta ,\gamma$ be three roots of a cubic equation $p{x^3} + q{x^2} + rx + s = 0$. Then, by the relation between roots and coefficients we get,
$\alpha + \beta + \gamma = \dfrac{{ - q}}{p} \\ \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p} \\ \alpha \beta \gamma = \dfrac{{ - s}}{p} \\$

It is given that the equation is ${x^3} - 12{x^2} + 39x - 28 = 0$.
We know that the highest power of any equation is the number of roots of that equation. Since, the highest power the given equation is three, so, the given equation has three roots.
It is also given that; the roots of the given equation are in A.P.
Let us consider the roots as $a - d,a,a + d$ where, $d$ is the common difference.
We have to find the value of this common difference that means the value of $d$.
Now, from the relation between roots and coefficient we get,
$a - d + a + a + d = 12$….. (1)
$a(a - d) + a(a + d) + (a - d)(a + d) = 39$… (2)
$(a - d)a(a + d) = 28$… (3)
Solving the equation (1) we get,
$3a = 12$
(We can eliminate $d$ as the sum of negative and positive value of same number is always zero)
Dividing we get,
$a = 4$
Next, we will find the value of $d$.
From equation (2) we get,
${a^2} - ad + {a^2} + ad + {a^2} - {d^2} = 39$
Eliminating $- ad$ and $ad$ we get,
$3{a^2} - {d^2} = 39$
Substitute the value of $a = 4$ in above equation we get,
$3{(4)^2} - {d^2} = 39$
Simplifying we get,
${d^2} = 48 - 39$
Simplifying again we get,
${d^2} = 9$
Taking square root of both side we get,
$d = \pm 3$
Hence,
The common difference of the roots of the given equation is $d = \pm 3$
The correct option is (C) $d = \pm 3$.
Here, we get two values of common difference that is $d = \pm 3$. Let us try to find out the roots.
For, $d = 3$, the roots are, $4 - 3,4,4 + 3$ that is $1,4,7$
Again,
For, $d = - 3$, the roots are, $4 + 3,4,4 - 3$ that is $7,4,1.$
It shows that for two different values of common difference, the roots will always be the same.

Note:An arithmetic progression is a sequence of numbers in which each term is derived from the preceding term by adding or subtracting a fixed number called the common difference "d".
The general form of an arithmetic progression is $a - d,{\text{ }}a,{\text{ }}a + d,{\text{ }}a + 2d,....$and so on. Thus the nth term of an AP series is ${T_n} = a + (n - 1)d$, where ${T_n} = {\text{nth }}$term and a is first term. Here d is a common difference = ${T_n} = {T_{n - 1}}$