If log ⁡2,log ⁡2n−1 and 2n+3 are in AP, then n is equal to

Since, 1og2 , log(2n -1) and log (2n+ 3) are in AP. ∴2log⁡2n−1=log⁡2+log⁡2n+3⇒2n−12=22n+3⇒2n−52n+1=0As 2n cannot be negative hence, 2n−5=0⇒ 2n=5⇒n=log2⁡5