Le=t $0<a_1<b_1$ For $n\geq 1$, define $a_{n+1}=\sqrt{a_n b_n}$ and $b_{n-1}=\frac{a_n+b_n}{2}.$ Then

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Let $0<a_1<b_1$ For $n\geq 1$, define $$a_{n+1}=\sqrt{a_n b_n}$$ and $$b_{n-1}=\frac{a_n+b_n}{2}.$$ Then which of the following is NOT TRUE?

(A) Both $\{a_n\}$ and $\{b_n\}$ converges, but the limits are not equal

(B) Both $\{a_n\}$ and $\{b_n\}$ converges, but the limits are equal

(C) $\{b_n\}$ is a decreasing sequence

(D) $\{a_n\}$ is an increasing sequence

My try $b_n-b_{n-1}=\frac{a_{n+1}+b_{n+1}}{2}-\frac{a_n+b_n}{2}=\frac{a_{n+1}-a_{n}}{2}+\frac{b_{n+1}-b_n}{2}$

From this, we get $\frac{a_{n+1}-a_{n}}{2}=b_n-b_{n-1}-\frac{b_{n+1}-b_n}{2}$

Using AM-GM inequality, $$a_{n+1}\leq b_{n-1}$$

$$a_3<b_1$$ $$a_4<b_2$$ $$a_5<b_3$$ $$a_6<b_4$$ $$...$$

I am not able to judge anything. Please help me.