The sixth term in the expansion of ${(3 - \sqrt{\frac{17}{4} + 3 \sqrt{2}})}^{10}$ is a

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negative irrational number

positive irrational number

positive rational number

negative rational number

answer is D.

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Detailed Solution

$\frac{\sqrt{17 + 12 \sqrt{2}}}{4} = \frac{\sqrt{17 + 2 \sqrt{72}}}{2} = \frac{3 + 2 \sqrt{2}}{2} G . B = {[3 - (\frac{3 + 2 \sqrt{2}}{2})]}^{10} = {(\frac{3 - 2 \sqrt{2}}{2})}^{10}$

$T_{6} = {}^{10}{C_{5}} \frac{{(3)}^{5} {(- 2 \sqrt{2})}^{5}}{2^{10}} =$ negative irrational

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