1. The problem is to verify the correctness of the given geometric sequence terms expressed as $a_n = a_m \cdot q$ for various indices $n$ and $m$. 2. Recall the formula for the $n$-th term of a geometric sequence: $$a_n = a_1 \cdot q^{n-1}$$ where $a_1$ is the first term and $q$ is the common ratio. 3. To check if $a_n = a_m \cdot q$ holds, substitute the formula: $$a_n = a_1 \cdot q^{n-1}$$ $$a_m = a_1 \cdot q^{m-1}$$ 4. Then, $$a_m \cdot q = (a_1 \cdot q^{m-1}) \cdot q = a_1 \cdot q^{m-1+1} = a_1 \cdot q^m$$ 5. For $a_n = a_m \cdot q$ to be true, we need: $$a_1 \cdot q^{n-1} = a_1 \cdot q^m$$ which implies $$q^{n-1} = q^m$$ 6. Since $q \neq 0$, this means: $$n-1 = m$$ 7. Now check each given expression: - $a_4 = a_2 \cdot q?$ Here, $n=4$, $m=2$, check if $4-1=2$? $3 \neq 2$, so false. - $a_5 = a_2 \cdot q?$ $5-1=4$, $4 \neq 2$, false. - $a_7 = a_3 \cdot q?$ $7-1=6$, $6 \neq 3$, false. - $a_8 = a_5 \cdot q?$ $8-1=7$, $7 \neq 5$, false. - $a_{15} = a_5 \cdot q?$ $15-1=14$, $14 \neq 5$, false. - $a_{30} = a_{10} \cdot q?$ $30-1=29$, $29 \neq 10$, false. 8. Conclusion: None of the given equalities $a_n = a_m \cdot q$ are correct because the condition $n-1 = m$ is not satisfied in any case. Final answer: All given expressions are false.