1. **Stating the problem:** We are given three consecutive terms of an exponential sequence: $$10 \cdot \binom{n}{6} / 7, \binom{b}{7}, \binom{n}{8}$$ We need to find the value of $n$ and the common ratio of the sequence. 2. **Understanding the problem:** In an exponential sequence, the ratio between consecutive terms is constant. Let the common ratio be $r$. Then: $$\frac{\binom{b}{7}}{10 \cdot \binom{n}{6} / 7} = r \quad \text{and} \quad \frac{\binom{n}{8}}{\binom{b}{7}} = r$$ 3. **Equating the ratios:** Since both ratios equal $r$, we have: $$\frac{\binom{b}{7}}{10 \cdot \binom{n}{6} / 7} = \frac{\binom{n}{8}}{\binom{b}{7}}$$ Cross-multiplying: $$\binom{b}{7}^2 = \frac{10}{7} \cdot \binom{n}{6} \cdot \binom{n}{8}$$ 4. **Simplify the binomial coefficients:** Recall the formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ 5. **Express $\binom{n}{6}$ and $\binom{n}{8}$:** $$\binom{n}{6} = \frac{n!}{6!(n-6)!}, \quad \binom{n}{8} = \frac{n!}{8!(n-8)!}$$ 6. **Multiply $\binom{n}{6} \cdot \binom{n}{8}$:** $$\binom{n}{6} \cdot \binom{n}{8} = \frac{(n!)^2}{6!8!(n-6)!(n-8)!}$$ 7. **Rewrite the equation:** $$\binom{b}{7}^2 = \frac{10}{7} \cdot \frac{(n!)^2}{6!8!(n-6)!(n-8)!}$$ 8. **Note that $b = n$ (since $b$ is not defined otherwise and terms involve $n$):** So, $$\binom{n}{7}^2 = \frac{10}{7} \cdot \frac{(n!)^2}{6!8!(n-6)!(n-8)!}$$ 9. **Express $\binom{n}{7}$:** $$\binom{n}{7} = \frac{n!}{7!(n-7)!}$$ 10. **Square $\binom{n}{7}$:** $$\binom{n}{7}^2 = \frac{(n!)^2}{(7!)^2 (n-7)!^2}$$ 11. **Substitute into the equation:** $$\frac{(n!)^2}{(7!)^2 (n-7)!^2} = \frac{10}{7} \cdot \frac{(n!)^2}{6!8!(n-6)!(n-8)!}$$ 12. **Cancel $(n!)^2$ from both sides:** $$\frac{1}{(7!)^2 (n-7)!^2} = \frac{10}{7} \cdot \frac{1}{6!8!(n-6)!(n-8)!}$$ 13. **Rewrite factorial terms:** Note that: $$(n-6)! = (n-6)(n-7)!$$ $$(n-8)! = (n-8)(n-9)!$$ But to simplify, consider the ratio: $$\frac{(n-6)!(n-8)!}{(n-7)!^2} = \frac{(n-6)(n-7)! (n-8)(n-9)!}{(n-7)!^2} = (n-6)(n-8) \cdot \frac{(n-9)!}{(n-7)!}$$ Since $(n-7)! = (n-7)(n-8)(n-9)!$, then: $$\frac{(n-9)!}{(n-7)!} = \frac{1}{(n-7)(n-8)}$$ So, $$\frac{(n-6)!(n-8)!}{(n-7)!^2} = (n-6)(n-8) \cdot \frac{1}{(n-7)(n-8)} = \frac{n-6}{n-7}$$ 14. **Substitute back:** $$\frac{1}{(7!)^2} \cdot \frac{1}{(n-7)!^2} = \frac{10}{7} \cdot \frac{1}{6!8!} \cdot \frac{1}{(n-6)!(n-8)!}$$ Using the ratio from step 13: $$\frac{1}{(7!)^2} \cdot \frac{1}{(n-7)!^2} = \frac{10}{7} \cdot \frac{1}{6!8!} \cdot \frac{1}{(n-7)!^2} \cdot \frac{n-7}{n-6}$$ 15. **Cancel $\frac{1}{(n-7)!^2}$ from both sides:** $$\frac{1}{(7!)^2} = \frac{10}{7} \cdot \frac{1}{6!8!} \cdot \frac{n-7}{n-6}$$ 16. **Calculate factorial values:** $$6! = 720, \quad 7! = 5040, \quad 8! = 40320$$ 17. **Substitute factorials:** $$\frac{1}{(5040)^2} = \frac{10}{7} \cdot \frac{1}{720 \times 40320} \cdot \frac{n-7}{n-6}$$ 18. **Calculate constants:** $$\frac{1}{(5040)^2} = \frac{1}{25401600}$$ $$\frac{10}{7} \cdot \frac{1}{720 \times 40320} = \frac{10}{7} \cdot \frac{1}{29030400} = \frac{10}{7 \times 29030400} = \frac{10}{203212800} = \frac{1}{20321280}$$ 19. **Rewrite equation:** $$\frac{1}{25401600} = \frac{1}{20321280} \cdot \frac{n-7}{n-6}$$ 20. **Multiply both sides by $20321280$:** $$\frac{20321280}{25401600} = \frac{n-7}{n-6}$$ 21. **Simplify fraction:** $$\frac{20321280}{25401600} = \frac{n-7}{n-6}$$ Divide numerator and denominator by 40320: $$\frac{503}{630} = \frac{n-7}{n-6}$$ 22. **Solve for $n$:** $$\frac{n-7}{n-6} = \frac{503}{630}$$ Cross-multiplied: $$630(n-7) = 503(n-6)$$ $$630n - 4410 = 503n - 3018$$ $$630n - 503n = 4410 - 3018$$ $$127n = 1392$$ $$n = \frac{1392}{127} = 10.96 \approx 11$$ 23. **Check $n=11$:** Calculate the common ratio $r$: $$r = \frac{\binom{11}{7}}{10 \cdot \binom{11}{6} / 7}$$ Calculate binomial coefficients: $$\binom{11}{6} = 462, \quad \binom{11}{7} = 330$$ 24. **Calculate $r$:** $$r = \frac{330}{10 \cdot 462 / 7} = \frac{330}{10 \cdot 66} = \frac{330}{660} = \frac{1}{2}$$ 25. **Final answer:** $$n = 11, \quad r = \frac{1}{2}$$