Let A_1, G_1, H_1 denote the arithmetic,geome | vedclass

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(C, A, B) $1.$ Given $A_n = \frac{A_{n-1}   H_{n-1}}{2}$,$G_n = \sqrt{A_{n-1} H_{n-1}}$,and $H_n = \frac{2 A_{n-1} H_{n-1}}{A_{n-1}   H_{n-1}}$.Note t

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$A, A, B$

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(C, A, B) $1.$ Given $A_n = \frac{A_{n-1} H_{n-1}}{2}$,$G_n = \sqrt{A_{n-1} H_{n-1}}$,and $H_n = \frac{2 A_{n-1} H_{n-1}}{A_{n-1} H_{n-1}}$.Note that $G_n^2 = A_{n-1} H_{n-1} = A_n H_n$. Also,$G_n = \sqrt{A_{n-1} H_{n-1}} = \sqrt{G_{n-1}^2} = G_{n-1}$.Thus,$G_1 = G_2 = G_3 = \ldots = \sqrt{ab}$.$2.$ Since $A_n$ is the arithmetic mean of $A_{n-1}$ and $H_{n-1}$,and $A_{n-1} > H_{n-1}$ for $n \geq 2$,we have $A_{n-1} > A_n > H_{n-1}$.Since $H_{n-1} A_2 > A_3 > \ldots$.$3.$ Since $H_n$ is the harmonic mean of $A_{n-1}$ and $H_{n-1}$,and $A_{n-1} > H_{n-1}$,we have $A_{n-1} > H_n > H_{n-1}$.Since $H_n > H_{n-1}$,it follows that $H_1 < H_2 < H_3 < \ldots$.

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