If the $A.M.$ between $p^{th}$ and $q^{th}$ terms of an $A.P.$ is equal to the $A.M.$ between $r^{th}$ and $s^{th}$ terms of the same $A.P.$,then $p + q$ is equal to

  • A
    $r + s - 1$
  • B
    $r + s - 2$
  • C
    $r + s + 1$
  • D
    $r + s$

Explore More

Similar Questions

Let $a_1, a_2, \ldots, a_n$ be in an $A$.$P$. If $a_5 = 2a_3$ and $a_{11} = 18$,then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}} + \ldots + \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.

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 $A.P.$

The sum of all two-digit positive numbers which,when divided by $7$,yield $2$ or $5$ as a remainder is:

If $a, b, c$ are in Arithmetic Progression,then $(a + 2b - c)(2b + c - a)(a + 2b + c) = \dots$

Statement-$I$: If the sum of $n$ terms of a sequence is $6n^2 + 3n + 1$,then it is an Arithmetic Progression $(AP)$.
Statement-$II$: The sum of $n$ terms of an Arithmetic Progression is always in the form $an^2 + bn$.

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