A) 55
B) 56
C) 45
D) 15
Correct Answer: A
Solution :
In the expansion of\[{{\left( 2+\frac{x}{3} \right)}^{n}},\] 7th term \[{{=}^{n}}{{C}_{7}}{{(2)}^{n-7}}{{\left( \frac{x}{3} \right)}^{7}}\] \[\therefore \]Coefficient of \[{{x}^{7}}{{=}^{n}}{{C}_{7}}{{(2)}^{n-7}}{{\left( \frac{1}{3} \right)}^{7}}\] and 8th term\[{{x}^{8}}{{=}^{n}}{{C}_{8}}{{(2)}^{n-8}}{{\left( \frac{x}{3} \right)}^{8}}\] \[\therefore \]Coefficient of \[{{x}^{8}}{{=}^{n}}{{C}_{8}}{{(2)}^{n-8}}{{\left( \frac{1}{3} \right)}^{8}}\] According to question \[^{n}{{C}_{7}}{{(2)}^{n-7}}{{\left( \frac{1}{3} \right)}^{7}}{{=}^{n}}{{C}_{8}}{{(2)}^{n-8}}{{\left( \frac{1}{3} \right)}^{8}}\] \[\Rightarrow \] \[^{n}{{C}_{7}}\times 2{{=}^{n}}{{C}_{8}}\times \frac{1}{3}\] \[\Rightarrow \] \[\frac{n!}{(n-7)!7!}\times 2=\frac{n!}{(n-8)!8!\times 3}\] \[\Rightarrow \] \[\frac{2}{(n-7)(n-8)!7!}=\frac{1}{(n-8)!8.7!.3}\] \[\Rightarrow \] \[\frac{2}{n-7}=\frac{1}{8\times 3}\] \[\Rightarrow \] \[n-7=48\] \[\Rightarrow \] \[n=48+7\] \[\Rightarrow \] \[n=55\]
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