B2. Which two consecutive terms of a geometric sequence with a common ratio of 3 must be multiplied to get 243 times the square of the first term of this sequence?
B2. Let $a_{1}$ be the first term. Then $a_{n}=a_{1} \cdot 3^{n-1}$ is the general term. The product of two consecutive terms is $a_{n} a_{n+1}=a_{1}^{2} \cdot 3^{2 n-1}$ and this is equal to $243 a_{1}^{2}$ when $3^{2 n-1}=243=3^{5}$, that is, at $n=3$. We need to multiply the third and fourth terms.