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

# If  and $\left({2}^{\mathrm{n}}+3\right)$ are in AP, then n is equal to

1. A

$\frac{5}{2}$

2. B

3. C

4. D

$\frac{3}{2}$

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### Solution:

Since, 1og2 , log(2n -1) and log (2n+ 3) are in AP.

$\begin{array}{ll}\therefore & 2\mathrm{log}\left({2}^{\mathrm{n}}-1\right)=\mathrm{log}2+\mathrm{log}\left({2}^{\mathrm{n}}+3\right)\\ ⇒& {\left({2}^{\mathrm{n}}-1\right)}^{2}=2\left({2}^{\mathrm{n}}+3\right)\\ ⇒& \left({2}^{\mathrm{n}}-5\right)\left({2}^{\mathrm{n}}+1\right)=0\end{array}$

As 2n cannot be negative hence,

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