If the sum of the roots of the equation $ax^{2} + bx + c = 0$ is equal to sum of the squares of their reciprocals, then $bc^{2}, ca^{2}, ab^{2}$ are in
- $A.P$
- $G.P$
- $H.P$
- $A.G.P$
Sum of roots = -b/a. Sum of squares of reciprocals = (1/r1)^2 + (1/r2)^2 = (r1^2 + r2^2) / (r1*r2)^2 = ((r1+r2)^2 - 2r1r2) / (r1r2)^2. Substituting r1+r2 = -b/a and r1r2 = c/a, we get (b^2/a^2 - 2c/a) / (c^2/a^2) = (b^2 - 2ac) / c^2. Equating -b/a = (b^2 - 2ac) / c^2 leads to the condition for AP.
Let the roots of the equation ax^2 + bx + c = 0 be alpha and beta. We are given that the sum of the roots equals the sum of the squares of their reciprocals, so alpha + beta = (1/alpha)^2 + (1/beta)^2. By Vieta's formulas, we replace the sum alpha + beta with -b/a and the product alpha*beta with c/a, yielding -b/a = (alpha^2 + beta^2)/(alpha^2*beta^2). Rewriting the numerator, we get -b/a = ((alpha+beta)^2 - 2*alpha*beta)/(alpha*beta)^2, which expands to -b/a = ((b^2/a^2) - 2c/a)/(c^2/a^2). Simplifying the numerator gives (b^2 - 2ac)/a^2, so the entire equation becomes -b/a = a(b^2 - 2ac)/c^2. Multiplying both sides by a*c^2 gives -bc^2 = a^2*b^2 - 2a^3*c, and rearranging the terms results in 2a^3*c = a^2*b^2 + bc^2. Dividing by the common factor abc leaves 2a^2/b = ab/c + c/a, proving that bc^2, ca^2, and ab^2 are in arithmetic progression.