Let $T_r$ be the $r$-th term of an arithmetic progression for $r = 1, 2, 3, \dots$. If for some positive integers $m$ and $n$,$T_m = \frac{1}{n}$ and $T_n = \frac{1}{m}$,then $T_{mn} = \dots$
- A$\frac{1}{mn}$
- B$\frac{1}{m} + \frac{1}{n}$
- C$1$
- D$0$
Explore More
Similar Questions
When the $9^{th}$ term of an $A.P.$ is divided by its $2^{nd}$ term,the quotient is $5$. When the $13^{th}$ term is divided by the $6^{th}$ term,the quotient is $2$ and the remainder is $5$. Find the first term of the $A.P.$
If $p$ times the $p^{th}$ term of an arithmetic progression is equal to $q$ times its $q^{th}$ term,then the $(p + q)^{th}$ term of this progression is........
Medium
View SolutionIf $a, b, c, d, e$ are in $A.P.$,then the value of $a + b + 4c - 4d + e$ in terms of $a$,if possible,is:
Easy
View SolutionIf $n$ is a natural number,then $\left( \frac{n+1}{2} \right)^n \ge n!$ is true when
Easy
View SolutionLet $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$ where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:
DifficultJEE MAIN 2013
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