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8. Sequences and Series
easy
If $\log 2,\;\log ({2^n} - 1)$ and $\log ({2^n} + 3)$ are in $A.P.$, then $n =$
A
$5/2$
B
${\log _2}5$
C
${\log _3}5$
D
$3/2$
Solution
(b) As, $\log 2,\;\log ({2^n} – 1)$ and $\log ({2^n} + 3)$ are in $A.P.$
Therefore, $2\log ({2^n} – 1) = \log 2 + \log ({2^n} + 3)$
$ \Rightarrow ({2^n} – 5)({2^n} + 1) = 0$
As ${2^n}$ cannot be negative, hence ${2^n} – 5 = 0$
$ \Rightarrow {2^n} = 5$ or $n = {\log _2}5$.
Standard 11
Mathematics
