The sum of $n$ A.M.'s between $a$ and $b$ is equal to

(a) $\frac{n}{2}(a+b)$
(b) $\frac{n+1}{2}(a+b)$
(c) $n + 1$
(d) $\frac{n+1}{2}$

Question

The sum of $n$ A.M.'s between $a$ and $b$ is equal to

(a) $\frac{n}{2}(a+b)$
(b) $\frac{n+1}{2}(a+b)$
(c) $n + 1$
(d) $\frac{n+1}{2}$

Accepted Answer

>

> [!TLDR]
> Sum of $n$ AMs $=\frac{n}{2}(a+b)$; option (a).
>
> Inserting $n$ arithmetic means between $a$ and $b$ gives an AP with $n+2$ terms. The $n$ means themselves have the same average as $a$ and $b$, namely $\frac{a+b}{2}$ (NCERT Class 11 Sequences a …

What you need	Formula
nth term	$T_n = a + (n-1)d$
Sum of n terms	$S_n = \frac{n}{2}[2a + (n-1)d]$
Sum using last term	$S_n = \frac{n}{2}(a + l)$
Common difference	$d = T_{n+1} - T_n$

Q.The sum of $n$ A.M.'s between $a$ and $b$ is equal to

Concept understanding — Arithmetic Progression

Arithmetic Progression: The Pattern of Equal Steps

The Core Idea

The General Term (nth term)

Why "Arithmetic"?

Sum of the First n Terms

Quick Reference

Q.The sum of nnn A.M.'s between aaa and bbb is equal to

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Q.The sum of $n$ A.M.'s between $a$ and $b$ is equal to

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