A) 18
B) 21
C) 28
D) 32
Correct Answer: A
Solution :
\[\frac{^{n}{{C}_{k}}}{^{n}{{C}_{k+1}}}=\frac{1}{2}\Rightarrow \frac{n!}{k!(n-k)!}\frac{(k+1)!(n-k-1)!}{n!}=\frac{1}{2}\] or\[\frac{k+1}{n-k}=\frac{1}{2}\] \[2k+2=n-k\] \[n-3k=2\] ?.(1) Similarly, \[\frac{^{n}{{C}_{k+1}}}{^{n}{{C}_{k+2}}}=\frac{2}{3}\] \[\frac{n!}{(k+1)!(n-k-1)!}.=\frac{(k+2)!(n-k-2)!}{n!}=\frac{2}{3}\] \[\frac{k+2}{n-k-1}=\frac{2}{3}\] \[3k+6=2n-2k-2\] \[2n-5k=8\] ?.(2) From (1) and (2) n = 14 and k = 4 \[\therefore \]\[\therefore \,\,n+k=18\]You need to login to perform this action.
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