MCQ
Given $a_1,a_2,a_3.....$ form an increasing geometric progression with common ratio $r$ such that $log_8a_1 + log_8a_2 +.....+ log_8a_{12} = 2014,$ then the number of ordered pairs of integers $(a_1, r)$ is equal to
- A$44$
- B$45$
- ✓$46$
- D$47$
✓
Answer
Correct option: C.
$46$
c
$\log _{8}\left(a_{1} a_{2} \dots a_{12}\right)=2014 \Rightarrow a_{1}^{12}. r^{66}=2^{6042} \dots(i)$
$\log _{8}\left(a_{1} a_{2} \dots a_{12}\right)=2014 \Rightarrow a_{1}^{12}. r^{66}=2^{6042} \dots(i)$
Let $a_{1}=2^{m}$ and $r=2^{n}$
$\therefore(1) \Rightarrow 2^{12 m+66 n}=2^{6042}$ ....$(i)$
$\Rightarrow 2 \mathrm{m}+11 \mathrm{n}=1007 \Rightarrow \mathrm{m}=\frac{1007-11 \mathrm{n}}{2} \in \mathrm{N}$
$\therefore$ allowed values of $n$ are $n=1,3,5,7, \ldots 91$
$\therefore$ total ordered pairs $=46$
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