Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
MediumIf $\frac{2+4+6+8+\dots+\text{upto } n \text{ terms}}{1+3+5+7+\dots+\text{upto } n \text{ terms}} = \frac{37}{36}$,then $n = $
- A$36$
- B$29$
- C$23$
- D$37$
Explore More
Similar Questions
The sum of the $1^{st} n$ terms of the series $\frac{1^{2}}{1} + \frac{1^{2}+2^{2}}{1+2} + \frac{1^{2}+2^{2}+3^{2}}{1+2+3} + \ldots$ is:
DifficultKCET 2016
View SolutionThe sum of the first $20$ terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is?
DifficultJEE MAIN 2018
View Solution$\sum\limits_{m = 1}^n {{m^2}}$ is equal to
Medium
View SolutionThe sum to $n$ terms of $(2n - 1) + 2(2n - 3) + 3(2n - 5) + \dots$ is
Medium
View SolutionThe $9^{th}$ term of the sequence $27, 9, 5\frac{2}{5}, 3\frac{6}{7}, \dots$ is $.....$
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo