Six positive numbers are in GP, such that the | vedclass

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(C) Let the six terms of the $GP$ be $\frac{a}{r^5}, \frac{a}{r^3}, \frac{a}{r}, ar, ar^3, ar^5$.Given that the product of these terms is $1000$:$\fra

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$\frac{1}{100}$

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(C) Let the six terms of the $GP$ be $\frac{a}{r^5}, \frac{a}{r^3}, \frac{a}{r}, ar, ar^3, ar^5$.Given that the product of these terms is $1000$:$\frac{a}{r^5} \cdot \frac{a}{r^3} \cdot \frac{a}{r} \cdot ar \cdot ar^3 \cdot ar^5 = 1000$$a^6 = 1000 = 10^3$$a^2 = (10^3)^{1/3} = 10$.Given the fourth term is $1$:$ar = 1 \Rightarrow a^2r^2 = 1$.Substituting $a^2 = 10$:$10r^2 = 1 \Rightarrow r^2 = \frac{1}{10}$.The last term is $ar^5 = \sqrt{a^2(r^2)^5} = \sqrt{10 \cdot (\frac{1}{10})^5} = \sqrt{\frac{1}{10^4}} = \frac{1}{100}$.

Six positive numbers are in $GP$, such that their product is $1000$. If the fourth term is $1$, then the last term is

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