6.1. The geometric progression $b_{1}, b_{2}, \ldots$ is such that $b_{25}=2 \operatorname{tg} \alpha, b_{31}=2 \sin \alpha$ for some acute angle $\alpha$. Find the number $n$ for which $b_{n}=\sin 2 \alpha$.

#
# Answer: 37

Solution. Let $q$ be the common ratio of our progression. Using the fact that $b_{31}=b_{25} q^{6}$ and $\sin 2 \alpha=2 \sin \alpha \cos \alpha$. We have $q^{6}=\frac{b_{31}}{b_{25}}=\frac{2 \sin \alpha}{2 \tan \alpha}=\cos \alpha$.

On the other hand, $\frac{\sin 2 \alpha}{2 \sin \alpha}=\cos \alpha$, from which $\sin 2 \alpha=b_{31} \cdot q^{6}=b_{37}$.

We see that $n=37$ fits. No other $n$ fits, since $0<q^{6}<1$, which means $0<|q|<1$, so no two terms of the progression coincide.