A) 108
B) 109
C) 110
D) 111
Correct Answer: D
Solution :
We have, a, b, c are in GP. |
\[\therefore \] \[{{b}^{2}}=ac\] |
and \[{{\log }_{6}}a+{{\log }_{6}}b+{{\log }_{6}}c=6\] \[\Rightarrow \]\[{{\log }_{6}}(abc)=6\]\[\Rightarrow \]\[abc={{6}^{6}}\] \[\Rightarrow \]\[{{b}^{3}}={{6}^{6}}\]\[\Rightarrow \]\[b=36\] \[\Rightarrow \]\[ac=36\times 36={{2}^{4}}\times {{3}^{4}}\] |
\[\Rightarrow \]\[b-a={{N}^{2}}\] \[\Rightarrow \]\[36-a={{N}^{2}}\] |
a is a factor of \[{{2}^{4}}\times {{3}^{4}}\] , \[a=27\]is possible value |
\[\therefore \]\[a=27,\]\[b=36,\]\[c=48\] |
\[a+b+c=27+36+48=111\] |
You need to login to perform this action.
You will be redirected in
3 sec