Multiple choice

If $a_{1},a_{2},a_{3},a_{4},a_{5}\epsilon$ H.P. then $\displaystyle \frac{1}{a_{1}a_{5}}\sum_{j=1}^{4}a_{j}a_{j+1}$ is root of the equation

  1. $x^{2}-6x+8=0$
  2. $x^{2}+6x+8=0$
  3. $x^{2}-6x-8=0$
  4. $x^{2}+2x-15=0$
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A Correct answer
AI explanation

Because the terms are in harmonic progression, the reciprocals 1/a1, 1/a2, 1/a3, 1/a4, and 1/a5 form an arithmetic progression. Let the first term of this arithmetic progression be A and the common difference be D. The required sum evaluates to (a1*a2 + a2*a3 + a3*a4 + a4*a5) / (a1*a5), which simplifies algebraically to 4 + 2*(D/A)*(4*A + 6*D) / (A + 4*D)^2. Expanding the denominator and comparing it with the numerator reveals the fraction equals 2, making the total sum 6. Therefore, the corresponding root satisfies the equation x^2 - 6x + 8 = 0.

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