2. Given a geometric sequence $\left\{a_{n}\right\}$ satisfies $a_{6}+a_{5}+a_{4}-a_{3}-a_{2}-a_{1}=49$, then the minimum value of $a_{9}+a_{8}+a_{7}$ is
Let the common ratio of $\left\{a_{n}\right\}$ be $\sqrt[3]{q}, a_{1}+a_{2}+a_{3}=a, a_{4}+a_{5}+a_{6}=a q, a_{7}+a_{8}+a_{9}=a q^{2}$, then $a q-a=49 \Rightarrow a=\frac{49}{q-1} \Rightarrow a q^{2}=\frac{49 q^{2}}{q-1}=49(q-1)+\frac{49}{q-1}+98$ $\geqslant 2 \cdot 49+98=196$, equality holds when $q=2$. Therefore, the minimum value of $a_{9}+a_{8}+a_{7}$ is 196.