Given a₁,a₂,a₃..... form an increasing geometric progression with common ratio r su

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8. Sequences and Series

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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

$45$

C

$46$

D

$47$

Solution

$\log _{8}\left(a_{1} a_{2} \dots a_{12}\right)=2014 \Rightarrow a_{1}^{12}. r^{66}=2^{6042} \dots(i)$

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 $=46$

Standard 11

Mathematics

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