Q.Solve $4x^3 - 24x^2 + 23x + 18 = 0$, given that the roots of this equation are in arithmetic progression.

Write the roots as $a-d, a, a+d$; the sum of roots gives $a$ directly, and the product of roots then gives $d$.

Let the roots of $4x^3-24x^2+23x+18=0$ be $a-d,\ a,\ a+d$ (in AP).

Sum of roots $= -\dfrac{\text{coeff. of }x^2}{\text{coeff. of }x^3} = -\dfrac{-24}{4}=6$:

$(a-d)+a+(a+d) = 3a = 6 \implies a=2$

Check: $a=2$ should be a root — substitute into the equation:

$4(2)^3-24(2)^2+23(2)+18 = 32-96+46+18 = 0 \ \checkmark$

Product of roots $= -\dfrac{\text{constant term}}{\text{coeff. of }x^3} = -\dfrac{18}{4}=-\dfrac92$:

$(a-d)(a)(a+d) = a(a^2-d^2) = -\frac92$

$2(4-d^2) = -\frac92 \implies 8-2d^2=-\frac92 \implies 2d^2 = 8+\frac92=\frac{25}2 \implies d^2=\frac{25}4 \implies d=\pm\frac52$

Taking $d=\dfrac52$ (the other sign just reverses the order): …