A) \[\frac{1}{2}abc\]
B) abc
C) 2 abc
D) 4 abc
Correct Answer: D
Solution :
\[(a+2b-c)\,(2b+c-a)\,(c+a-b)\] \[=(a+a+c-c)(a+c+c-a)(2b-b)\]\[=4\,abc.\] \[(\because a,b,c\] are in A.P., \[\therefore 2b=a+c)\].
You need to login to perform this action.
You will be redirected in
3 sec
