A) \[2\,(c-a)\]
B) \[2\,(d-b)\]
C) \[2\,(f-d)\]
D) \[2(d-c)\]
Correct Answer: D
Solution :
\[\because \] a, b, c, d, e f are in AP. So, \[b-a=c-b=d-c=e-d=f-e\] Now, \[d-c=e-d\] \[\Rightarrow \] \[e+c=2d\] \[\Rightarrow \] \[e-c+2c=2d\] \[\Rightarrow \] \[e-c=2(d-c)\]You need to login to perform this action.
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