(D) The $n^{th}$ term of a $G.P.$ is given by $l = a r^{n-1}$.Dividing both sides by $a$,we get $\frac{l}{a} = r^{n-1}$.Taking the logarithm on both s

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$1 + \frac{\log l - \log a}{\log r}$

(D) The $n^{th}$ term of a $G.P.$ is given by $l = a r^{n-1}$.Dividing both sides by $a$,we get $\frac{l}{a} = r^{n-1}$.Taking the logarithm on both sides,we have $\log(\frac{l}{a}) = \log(r^{n-1})$.Using the properties of logarithms,$\log l - \log a = (n-1) \log r$.Therefore,$n-1 = \frac{\log l - \log a}{\log r}$.Adding $1$ to both sides,we get $n = 1 + \frac{\log l - \log a}{\log r}$.

Class 11 Mathematics Sequences and Series Geometric progression

Easy

The first and last terms of a $G.P.$ are $a$ and $l$ respectively; $r$ being its common ratio; then the number of terms in this $G.P.$ is

A
$\frac{\log l - \log a}{\log r}$
B
$1 - \frac{\log l - \log a}{\log r}$
C
$\frac{\log a - \log l}{\log r}$
D
$1 + \frac{\log l - \log a}{\log r}$

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Class 11 Questions

Class 11

Mathematics Questions

Chapter

Sequences and Series

Subtopic

Geometric progression

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$ such that $B_n > A_n$ for all $n \geq p$ is:

DifficultJEE MAIN 2018

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The $5^{\text{th}}$,$8^{\text{th}}$,and $11^{\text{th}}$ terms of a $G.P.$ are $p, q$,and $s$,respectively. Show that $q^{2} = ps$.

Medium

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If the ratio of the sum of the first three terms and the sum of the first six terms of a $G.P.$ is $125 : 152$,then the common ratio $r$ is:

Easy

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If $n$ geometric means between $a$ and $b$ are $G_1, G_2, ..., G_n$ and a single geometric mean is $G$,then the true relation is

Difficult

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Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive terms. If $a_1 a_5 = 28$ and $a_2 + a_4 = 29$,then $a_6$ is equal to

DifficultJEE MAIN 2025

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The first and last terms of a $G.P.$ are $a$ and $l$ respectively; $r$ being its common ratio; then the number of terms in this $G.P.$ is

Similar Questions

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$ such that $B_n > A_n$ for all $n \geq p$ is:

The $5^{\text{th}}$,$8^{\text{th}}$,and $11^{\text{th}}$ terms of a $G.P.$ are $p, q$,and $s$,respectively. Show that $q^{2} = ps$.

If the ratio of the sum of the first three terms and the sum of the first six terms of a $G.P.$ is $125 : 152$,then the common ratio $r$ is:

If $n$ geometric means between $a$ and $b$ are $G_1, G_2, ..., G_n$ and a single geometric mean is $G$,then the true relation is

Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive terms. If $a_1 a_5 = 28$ and $a_2 + a_4 = 29$,then $a_6$ is equal to

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