Multiple choice

Lets consider quadratic equation $ax^2 + bx + c = 0$ where $a, b, c \in R$ and $a\ne 0$. If above equation has roots $\alpha, \beta$, then $\displaystyle{\alpha + \beta = -\frac{b}{a} , \alpha\beta = \frac{c}{a}}$ and the equation can be written as $ax^2+ bx + c = a(x - \alpha)(x - \beta)$. Also, if $a_1, a_2, a_3, a_4$, ..... are in A.P., then $a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = ... \ne 0$ and if $b_1, b_2, b_3, b_4$, ... are in G.P., then $\displaystyle{\frac{b_2}{b_1} = \frac{b_3}{b_2} = \frac{b_4}{b_3}} =$ ... $\ne1$ Now if $c_1$, $c_2$, $c_3$, $c_4$, ... are in HP, then $\displaystyle{\frac{1}{c_2} - \frac{1}{c_1} = \frac{1}{c_3} - \frac{1}{c_2} = \frac{1}{c_4} - \frac{1}{c_3}} = $... $\ne 0$. If the roots of equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ are equal, then $a, b, c$ are in

  1. $A.P.$
  2. $G.P.$
  3. $H.P.$
  4. $\text {None of these}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the equation a(b - c)x^2 + b(c - a)x + c(a - b) = 0, the sum of coefficients is a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0. Thus, x = 1 is a root. Since the roots are equal, the other root is also 1. The product of roots is c(a-b) / a(b-c) = 1, so c(a-b) = a(b-c), which leads to 2ac = ab + bc, or 2/b = 1/c + 1/a, meaning a, b, c are in HP.

AI explanation

For the equation a(b-c)x^2 + b(c-a)x + c(a-b) = 0, you can easily verify by substitution that x = 1 is always a root. Since the problem states the roots are equal, the other root must also be 1, meaning the product of the roots is 1. Using the product of the roots formula, c(a-b) divided by a(b-c) equals 1, which simplifies to ac minus bc equals ab minus ac. Rearranging this equation gives b equals 2ac divided by (a+c), which is the exact condition that proves a, b, and c are in Harmonic Progression.