A) \[n-m+1:m\]
B) \[n-m+1:n\]
C) \[m:n-m+1\]
D) \[n:n-m+1\]
Correct Answer: C
Solution :
Since, n arithmetic means are inserted between a and 2b, \[\therefore \] \[{{a}_{m}}\] (mth mean) \[=a+\frac{m}{n+1}(2b-a)\] and n arithmetic means are inserted between 2a and b. \[\therefore \]\[a{{}_{m}}\](mth mean) \[=2a+\frac{m}{n+1}(b-2a)\] Given, \[{{a}_{m}}=a{{}_{m}}\] \[\Rightarrow \]\[a+\frac{m}{n+1}.(2b-a)=2a+\frac{m}{(n+1)}(b-2a)\] \[\Rightarrow \] \[\frac{a}{b}=\frac{m}{n-m+1}\] \[\therefore \] \[a:b=m:n-m+1\]
You need to login to perform this action.
You will be redirected in
3 sec
