If a,4,bare in AP and a,2,bare in GP then 1a,1,1b are in
AP
GP
HP
none of these
Explanation for the correct option:
Step 1. Evaluating the given conditions:
∵a,4,b are in AP
⇒8=a+b …(1) ∵2b=a+b
∵a,2,bare in GP
⇒4=ab …(2) ∵b2=ab
Step 2. Divide equation (1) by (2), we get
⇒ (a+b)ab=84
⇒1b+1a=2
∴1a,1,1bare in AP
Hence, option A. is the correct answer:
Consider two events A and B such that P(A)=14, P(BA)=12, P(AB)=14. For each of the following statements, which is true.
I.P(A'B')=34
II. The events A and B are mutually exclusive
III.P(AB)+P(AB')=1