(A) The given $G.P.$ is $x^{3}, x^{5}, x^{7}, \ldots$ Here,the first term $a = x^{3}$ and the common ratio $r = \frac{x^{5}}{x^{3}} = x^{2}$. The sum

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$\frac{x^{3}(1-x^{2n})}{1-x^{2}}$

(A) The given $G.P.$ is $x^{3}, x^{5}, x^{7}, \ldots$ Here,the first term $a = x^{3}$ and the common ratio $r = \frac{x^{5}}{x^{3}} = x^{2}$. The sum of $n$ terms of a $G.P.$ is given by the formula $S_{n} = \frac{a(1-r^{n})}{1-r}$. Substituting the values of $a$ and $r$: $S_{n} = \frac{x^{3}(1-(x^{2})^{n})}{1-x^{2}}$ $S_{n} = \frac{x^{3}(1-x^{2n})}{1-x^{2}}$

Class 11 Mathematics Sequences and Series Geometric progression

Easy

Find the sum to the indicated number of terms in the geometric progression: $x^{3}, x^{5}, x^{7}, \ldots$ for $n$ terms (where $x \neq \pm 1$).

A
$\frac{x^{3}(1-x^{2n})}{1-x^{2}}$
B
$\frac{x^{3}(1-x^{n})}{1-x^{2}}$
C
$\frac{x^{2}(1-x^{2n})}{1-x^{2}}$
D
$\frac{x^{3}(1-x^{2n})}{1-x}$

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Sequences and Series

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Geometric progression

If the $5^{th}$ term of a geometric progression is $2$,what is the product of its first $9$ terms?

Easy

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If $|\alpha| < 1$ and $|\beta| < 1$,and $1 - \alpha + \alpha^2 - \alpha^3 + \dots \infty = s_1$ and $1 - \beta + \beta^2 - \beta^3 + \dots \infty = s_2$,then $1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$ equals:

Medium

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For $0 < \phi < \frac{\pi}{2}$,if $x = \sum_{n=0}^\infty \cos^{2n}\phi$,$y = \sum_{n=0}^\infty \sin^{2n}\phi$,and $z = \sum_{n=0}^\infty \cos^{2n}\phi \sin^{2n}\phi$,then:

DifficultIIT 1993

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If the roots of the cubic equation $ax^3 + bx^2 + cx + d = 0$ are in $G.P.$,then

Difficult

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If $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression and $\frac{a_3}{a_1}=25$,then $\frac{a_9}{a_5}$ equals

EasyKCET 2022

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Find the sum to the indicated number of terms in the geometric progression: $x^{3}, x^{5}, x^{7}, \ldots$ for $n$ terms (where $x \neq \pm 1$).

Similar Questions

If the $5^{th}$ term of a geometric progression is $2$,what is the product of its first $9$ terms?

If $|\alpha| < 1$ and $|\beta| < 1$,and $1 - \alpha + \alpha^2 - \alpha^3 + \dots \infty = s_1$ and $1 - \beta + \beta^2 - \beta^3 + \dots \infty = s_2$,then $1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$ equals:

For $0 < \phi < \frac{\pi}{2}$,if $x = \sum_{n=0}^\infty \cos^{2n}\phi$,$y = \sum_{n=0}^\infty \sin^{2n}\phi$,and $z = \sum_{n=0}^\infty \cos^{2n}\phi \sin^{2n}\phi$,then:

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If $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression and $\frac{a_3}{a_1}=25$,then $\frac{a_9}{a_5}$ equals

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