Given that (10)/(7) n6, n7, and n8 are the first three consecutive terms of a geometric sequence (G.P.).

Here are the solutions to question 9:

Given that $\frac{10}{7} \binom{n}{6}$, $\binom{n}{7}$, and $\binom{n}{8}$ are the first three consecutive terms of a geometric sequence (G.P.).

Let the terms be $a_1 = \frac{10}{7} \binom{n}{6}$, $a_2 = \binom{n}{7}$, and $a_3 = \binom{n}{8}$.
For a geometric progression, the common ratio $r$ is constant, so $\frac{a_2}{a_1} = \frac{a_3}{a_2}$.

We use the property of combinations: $\binom{n}{k} = \frac