Let A_{1}, A_{2}, A_{3}, ldots be an increasi | vedclass

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

Question

(C) Let the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 1$.Given $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$.Substituting the

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(C) Let the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 1$.Given $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$.Substituting the terms: $a \cdot (ar^2) \cdot (ar^4) \cdot (ar^6) = a^4 r^{12} = (ar^3)^4 = (A_{4})^4 = \frac{1}{1296}$.Thus,$A_{4} = \sqrt[4]{\frac{1}{1296}} = \frac{1}{6}$.Given $A_{2} + A_{4} = \frac{7}{36}$,we have $ar + ar^3 = \frac{7}{36}$.Since $A_{4} = ar^3 = \frac{1}{6}$,we have $ar + \frac{1}{6} = \frac{7}{36}$,so $ar = \frac{7}{36} - \frac{6}{36} = \frac{1}{36}$.Now,$r^2 = \frac{ar^3}{ar} = \frac{1/6}{1/36} = 6$,so $r = \sqrt{6}$.Then $a = \frac{ar}{r} = \frac{1/36}{\sqrt{6}} = \frac{1}{36\sqrt{6}}$.We need to find $A_{6} + A_{8} + A_{10} = ar^5 + ar^7 + ar^9 = ar^5(1 + r^2 + r^4)$.$ar^5 = (ar) \cdot r^4 = \frac{1}{36} \cdot (6)^2 = \frac{36}{36} = 1$.$A_{6} + A_{8} + A_{10} = 1 \cdot (1 + 6 + 36) = 43$.

Let $A_{1}, A_{2}, A_{3}, \ldots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$,then the value of $A_{6} + A_{8} + A_{10}$ is equal to

Similar Questions

The sum of the first two terms of a geometric progression is $12$. The sum of the third and fourth terms is $48$. The terms of the geometric progression alternate between positive and negative. What is the first term?

The three sides of a right-angled triangle are in $GP$ (geometric progression). If the two acute angles are $\alpha$ and $\beta$,then $\tan \alpha$ and $\tan \beta$ are

If the sum of the second,third and fourth terms of a positive term $G.P.$ is $3$ and the sum of its sixth,seventh and eighth terms is $243,$ then the sum of the first $50$ terms of this $G.P.$ is

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:

If $n$ geometric means are inserted between $a$ and $b$,then the $n^{th}$ geometric mean will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module