Given a_1,a_2,a_3..... form an increasing&nbs | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\log _{8}\left(a_{1} a_{2} \dots a_{12}ight)=2014 \Rightarrow a_{1}^{12}. r^{66}=2^{6042} \dots(i)$

Let $a_{1}=2^{m}$ and $r=2^{n}$

$	herefo

Vedclass · Accepted Answer

$46$

Vedclass · Answer

$\log _{8}\left(a_{1} a_{2} \dots a_{12}ight)=2014 \Rightarrow a_{1}^{12}. r^{66}=2^{6042} \dots(i)$

Let $a_{1}=2^{m}$ and $r=2^{n}$

$	herefore(1) \Rightarrow 2^{12 m+66 n}=2^{6042}$     ....$(i)$

$\Rightarrow 2 \mathrm{m}+11 \mathrm{n}=1007 \Rightarrow \mathrm{m}=\frac{1007-11 \mathrm{n}}{2} \in \mathrm{N}$

$	herefore$ allowed values of $n$ are $n=1,3,5,7, \ldots 91$

$	herefore$ total ordered pairs $=46$

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

Similar Questions

The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is

$x = 1 + a + {a^2} + ....\infty \,(a < 1)$ $y = 1 + b + {b^2}.......\infty \,(b < 1)$ Then the value of $1 + ab + {a^2}{b^2} + ..........\infty $ is

The ${20^{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + .......$ will be

If $1\, + \,\sin x\, + \,{\sin ^2}x\, + \,...\infty \, = \,4\, + \,2\sqrt 3 ,\,0\, < \,x\, < \,\pi $ then

If the product of three consecutive terms of $G.P.$ is $216$ and the sum of product of pair-wise is $156$, then the numbers will be