Given a_1, a_2, a_3, dots form an increasing | vedclass

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

Question

(C) Given $\log_8(a_1 a_2 \dots a_{12}) = 2014$. Since $a_k = a_1 r^{k-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$. Thus,$\log_

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(C) Given $\log_8(a_1 a_2 \dots a_{12}) = 2014$. Since $a_k = a_1 r^{k-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$. Thus,$\log_8(a_1^{12} r^{66}) = 2014 \Rightarrow a_1^{12} r^{66} = 8^{2014} = 2^{6042}$. Let $a_1 = 2^m$ and $r = 2^n$ where $m, n$ are integers. Then $(2^m)^{12} (2^n)^{66} = 2^{12m + 66n} = 2^{6042}$. This implies $12m + 66n = 6042$,which simplifies to $2m + 11n = 1007$. Since $r > 1$ and the progression is increasing,$n \ge 1$. Also $a_1 \ge 1$,so $m \ge 0$. $2m = 1007 - 11n$. For $m$ to be an integer,$1007 - 11n$ must be even,so $n$ must be odd. Also $m > 0$ $\Rightarrow 1007 - 11n > 0$ $\Rightarrow n Thus $n \in \{1, 3, 5, \dots, 91\}$. The number of such odd values is $\frac{91 - 1}{2} + 1 = 46$.

Given $a_1, a_2, a_3, \dots$ form an increasing geometric progression with common ratio $r$ such that $\log_8 a_1 + \log_8 a_2 + \dots + \log_8 a_{12} = 2014$,then the number of ordered pairs of integers $(a_1, r)$ is equal to

Similar Questions

If the roots of $x^3-42x^2+336x-512=0$ are in increasing geometric progression,then its common ratio is: (in $:1$)

Let the positive numbers $a_1, a_2, a_3, a_4$ and $a_5$ be in a $G$.$P$. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively,where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3+a_4+a_5=14$,then $m+n$ is equal to $.........$.

Consider a geometric progression with first term $a$ and common ratio $r$. If $A$ and $H$ are the arithmetic mean and harmonic mean of the first $n$ terms of the geometric progression respectively,then $A \cdot H = \dots$

The geometric mean of $n$ positive terms $x_1, x_2, \dots, x_n$ is equal to:

Find the sum of the first $9$ terms of the geometric series $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module