Multiple choice

If the sum of the roots of the quadratic equation $\displaystyle ax^{2}+bx+c=0 $ is equal to sum of the squares of their reciprocals then $\displaystyle \frac{a}{c},\frac{b}{a},\frac{c}{b} $ are in

  1. AP

  2. GP

  3. HP

  4. none of these
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C Correct answer
Explanation

Sum of roots = -b/a. Sum of squares of reciprocals = (b^2 - 2ac)/c^2. Equating: -b/a = (b^2 - 2ac)/c^2. This leads to -bc^2 = ab^2 - 2a^2c. Dividing by abc: -c/a = b/c - 2a/b. Rearranging: 2a/b = b/c + c/a. This is the condition for HP.

AI explanation

Let the roots be p and q, so their sum is p + q = -b/a and their product is pq = c/a. The condition states p + q equals (1/p)^2 + (1/q)^2, which expands to (p^2 + q^2)/(p^2 * q^2). Substituting the sum and product values gives (-b/a) = ((b^2 - 2ac)/a^2) / (c^2/a^2). This simplifies to -b/a = (b^2 - 2ac)/c^2, and cross-multiplying yields -bc^2 = ab^2 - 2a^2c. Dividing by abc results in -c/a = b/c - 2a/b, which rearranges to 2a/b = b/c + c/a. This proves the terms a/b, b/c, and c/a are in arithmetic progression, making their reciprocals b/a, c/b, and a/c in arithmetic progression; consequently, the inverted sequence a/c, b/a, c/b is in harmonic progression.

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