If reciprocals of the roots of equation $10x^{3}-cx^{2}-54x-27=0$ are in arithmetic progression, then value of $c$ is

Let the roots of the cubic equation be alpha, beta, and gamma, making their reciprocals 1 over alpha, 1 over beta, and 1 over gamma form an arithmetic progression. By Vieta's formulas for the original cubic, the sum of the roots gives alpha plus beta plus gamma equals c divided by 10, and the product of the roots gives alpha times beta times gamma equals 27 divided by 10. For three terms in an arithmetic progression, the sum is three times the middle term, so 1 over beta equals (1 over alpha plus 1 over beta plus 1 over gamma) divided by 3. Using Vieta's sum of products taken two at a time, (1 over alpha plus 1 over beta plus 1 over gamma) equals negative 54 divided by 27, which equals negative 2. Thus, the middle reciprocal 1 over beta equals negative 2 divided by 3, making beta equal negative 1.5. Substituting beta into the original equation gives 10(-1.5) cubed minus c(-1.5) squared minus 54(-1.5) minus 27 equals 0, which simplifies to negative 33.75 minus 2.25c plus 81 minus 27 equals 0. Solving this linear equation gives 2.25c equals 20.25, so c equals 9.

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