(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

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

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

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

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Sequences and Series

Subtopic

Geometric progression

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2\sum_{n=1}^{40}(P_{n})^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}$ and $\gcd(\alpha,\beta)=1$,then $\alpha+\beta$ is equal to

DifficultJEE MAIN 2026

View Solution

If $a, b$ and $c$ form a geometric progression with common ratio $r$,then the sum of the ordinates of the points of intersection of the line $ax + by + c = 0$ and the curve $x + 2y^2 = 0$ is

DifficultTS EAMCET 2012

View Solution

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

MediumWBJEE 2013

View Solution

The $7^{th}$ term of the sequence $\sqrt{2}, \sqrt{10}, 5\sqrt{2}, \dots$ is

Easy

View Solution

The sum of the first four terms of a geometric progression $(G.P.)$ is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$. If the product of the first three terms of the $G.P.$ is $1$,and the third term is $\alpha$,then $2\alpha$ is ....... .

DifficultJEE MAIN 2021

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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 $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2\sum_{n=1}^{40}(P_{n})^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}$ and $\gcd(\alpha,\beta)=1$,then $\alpha+\beta$ is equal to

If $a, b$ and $c$ form a geometric progression with common ratio $r$,then the sum of the ordinates of the points of intersection of the line $ax + by + c = 0$ and the curve $x + 2y^2 = 0$ is

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

The $7^{th}$ term of the sequence $\sqrt{2}, \sqrt{10}, 5\sqrt{2}, \dots$ is

The sum of the first four terms of a geometric progression $(G.P.)$ is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$. If the product of the first three terms of the $G.P.$ is $1$,and the third term is $\alpha$,then $2\alpha$ is ....... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module