A) \[2(c-a)\]
B) \[2(f-d)\]
C) \[2(d-c)\]
D) \[d-c\]
Correct Answer: C
Solution :
\[a,\ b,\ c,\ d,\ e,\ f\] are in A.P. So \[b-a=c-b=d-c=e-d=f-e=K\] Where \[K\] is a common difference. Now, \[d-c=e-d\]\[\Rightarrow \]\[e+c=2d\]. \[e\text{--}c+\text{2}c=2d\Rightarrow e-c=2(d-c)\]. Trick: Check by putting \[a=1,\ b=2,\ c=3,\ d=4,\ e=5\] and\[f=6\].
You need to login to perform this action.
You will be redirected in
3 sec
