6.16 The sum of three consecutive terms of a geometric progression is 65, and the sum of their logarithms to the base 15 is 3. Find these terms of the progression.

## Group B
6.16 Due to the condition, we have the system

$$
\left\{\begin{array}{l}
a_{1}+a_{1} q+a_{1} q^{2}=65 \\
\log _{15} a_{1}+\log _{15}\left(a_{1} q\right)+\log _{15}\left(a_{1} q^{2}\right)=3
\end{array}\right.
$$

Writing the second equation as $\log _{15}\left(a_{1} \cdot a_{1} q \cdot a_{1} q^{2}\right)=3$, we get $a_{1}^{3} q^{3}=15^{3}$, i.e., $a_{1} q=15$. Thus, we arrive at the system

$$
\left\{\begin{array} { l } 
{ a _ { 1 } q = 1 5 , } \\
{ a _ { 1 } + 1 5 + 1 5 q = 6 5 , }
\end{array} \text { or } \left\{\begin{array}{l}
a_{1} q=15 \\
a_{1}+15 q=50
\end{array}\right.\right.
$$

and as a result, we get two solutions:

1) $\left.a_{1}=5, q=3 ; 2\right) a_{1}=45, q=\frac{1}{3}$.

Answer: $5 ; 15 ; 45$ or $45 ; 15 ; 5$.