A) 5
B) 10
C) 15
D) 20
Correct Answer: D
Solution :
Given a, b, c are in GP and 4a, 5b, 4c are in AP \[\therefore \] \[{{b}^{2}}=ac\] and \[5b=\frac{4a+4c}{2}\] \[\Rightarrow \] \[{{b}^{2}}=ac\] and \[5b=2a+2c\] Now, \[a+b+c=70\] (Given) \[\Rightarrow \] \[2a+2c+2b=140\] \[\Rightarrow \] \[5b+2b=140\] \[\Rightarrow \] \[b=20\]You need to login to perform this action.
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