Six positive numbers are in GP,such that thei | vedclass

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

Question

(C) Let the six terms of the $GP$ be $\frac{a}{r^5}, \frac{a}{r^3}, \frac{a}{r}, ar, ar^3, ar^5$.Given that the product of these terms is $1000$:$\fra

Vedclass · Accepted Answer

$\frac{1}{100}$

Vedclass · Answer

(C) Let the six terms of the $GP$ be $\frac{a}{r^5}, \frac{a}{r^3}, \frac{a}{r}, ar, ar^3, ar^5$.Given that the product of these terms is $1000$:$\frac{a}{r^5} \cdot \frac{a}{r^3} \cdot \frac{a}{r} \cdot ar \cdot ar^3 \cdot ar^5 = 1000$$a^6 = 1000 = 10^3$$a^2 = (10^3)^{1/3} = 10$.Given the fourth term is $1$:$ar = 1 \Rightarrow a^2r^2 = 1$.Substituting $a^2 = 10$:$10r^2 = 1 \Rightarrow r^2 = \frac{1}{10}$.The last term is $ar^5 = \sqrt{a^2(r^2)^5} = \sqrt{10 \cdot (\frac{1}{10})^5} = \sqrt{\frac{1}{10^4}} = \frac{1}{100}$.

Six positive numbers are in $GP$,such that their product is $1000$. If the fourth term is $1$,then the last term is

Similar Questions

The $4^{th}$ term of a $G.P.$ is the square of its second term,and the first term is $-3$. Determine its $7^{th}$ term.

In an increasing geometric progression of positive terms,the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text{th}}$,$6^{\text{th}}$,and $8^{\text{th}}$ terms is:

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

In a geometric progression,if the ratio of the sum of the first $5$ terms to the sum of their reciprocals is $49$,and the sum of the first and the third term is $35$,then the first term of this geometric progression is:

Find the sum of the products of the corresponding terms of the sequences $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \frac{1}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module