Multiple choice

If the roots of the equation $x^{5}-80x^{4}+px^{3}+qx^{2}+rx-s=0$ are in G. P and all the terms are positive, such that the sum of its reciprocals is $20$, then find the value of s

  1. $8$
  2. $16$
  3. $32$
  4. $64$
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C Correct answer
Explanation

For a polynomial with roots in GP, the product of roots is s = a^5 * r^10. The sum of reciprocals is (1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5) = 20. For a 5th degree polynomial, this relates to the coefficients. Given the symmetry and properties of GP, s = 32.

AI explanation

Let the five positive roots in geometric progression be a divided by r squared, a divided by r, a, ar and ar squared. The sum of the roots gives 80 equals a multiplied by the sum of 1 divided by r squared, 1 divided by r, 1, r and r squared. The sum of the reciprocals of the roots is 20, which gives 20 equals 1 divided by a multiplied by the sum of r squared, r, 1, 1 divided by r and 1 divided by r squared. The sums in both equations are identical, so dividing the first equation by the second gives 4 equals a squared. Since the roots are positive, a equals 2. The product of the roots of a polynomial equals the constant term divided by the leading coefficient times negative one raised to the degree. For this quintic, the product equals s. Multiplying the roots gives a to the fifth power, which is 2 to the fifth power. Thus s equals 32.

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