7.047. Find the natural number $n$ from the equation

$$
3^{2} \cdot 3^{5} \cdot 3^{8} \cdots 3^{3 n-1}=27^{5}
$$
## Solution.

$3^{2+5+8+\ldots+3 n-1}=3^{15}, 2+5+8+\ldots+3 n-1=15$.

On the left side of the equation, we have the sum of the terms of an arithmetic progression $S_{k}$, where $a_{1}=2, d=3, a_{k}=3 n-1, k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-1-2}{3}+1=n$.

Then $S_{k}=\frac{a_{1}+a_{k}}{2} \cdot k=\frac{2+3 n-1}{2} \cdot n=\frac{3 n^{2}+n}{2}$, and the equation becomes $\frac{3 n^{2}+n}{2}=15,3 n^{2}+n-30=0$, from which $n=3$.

Answer: 3.

Solve the equations (7.048 - 7.127):