If A is an arithmetic mean between two number | vedclass

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

Question

(A) Let the two numbers be $a$ and $b$. The arithmetic mean $A$ between $a$ and $b$ is given by $A = \frac{a+b}{2}$. Let $A_1, A_2, \dots, A_n$ be the

Vedclass · Accepted Answer

$S = nA$

Vedclass · Answer

(A) Let the two numbers be $a$ and $b$. The arithmetic mean $A$ between $a$ and $b$ is given by $A = \frac{a+b}{2}$. Let $A_1, A_2, \dots, A_n$ be the $n$ arithmetic means between $a$ and $b$. Then $a, A_1, A_2, \dots, A_n, b$ form an arithmetic progression with $n+2$ terms. The sum of these $n$ arithmetic means is $S = A_1 + A_2 + \dots + A_n$. In an arithmetic progression,the sum of $n$ arithmetic means inserted between $a$ and $b$ is equal to $n$ times the single arithmetic mean between $a$ and $b$. Mathematically,$S = \sum_{i=1}^{n} A_i = n \left( \frac{a+b}{2} \right) = nA$. Therefore,$S = nA$.

If $A$ is an arithmetic mean between two numbers and $S$ is the sum of $n$ arithmetic means between the same two numbers,then:

Similar Questions

The sum of the first $20$ terms common between the series $3 + 7 + 11 + 15 + \dots$ and $1 + 6 + 11 + 16 + \dots$ is

Five numbers are in $A.P.$,whose sum is $25$ and product is $2520$. If one of these five numbers is $-\frac{1}{2}$,then the greatest number amongst them is

Let $A_1, A_2, \dots, A_{39}$ be $39$ arithmetic means between the numbers $59$ and $159$. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to:

If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in an arithmetic progression,then $x = \dots$

$a_1, a_2, a_3, \dots, a_{100}$ are in an arithmetic progression,where $a_1 = 3$ and $S_p = \sum_{i=1}^p a_i, 1 \le p \le 100$. For any integer $n$,let $m = 5n$. If $S_m/S_n$ is independent of $n$,then $a_2 = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module