If {A_1}, {A_2}; {G_1}, {G_2} and {H_1}, {H_2 | vedclass

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

Question

(A) Let the two quantities be $a$ and $b$.Since ${A_1}, {A_2}$ are $AMs$ between $a$ and $b$,$a, {A_1}, {A_2}, b$ are in $A.P.$Thus,${A_1} - a = b - {

Vedclass · Accepted Answer

$\frac{{A_1 + A_2}}{{H_1 + H_2}}$

Vedclass · Answer

(A) Let the two quantities be $a$ and $b$.Since ${A_1}, {A_2}$ are $AMs$ between $a$ and $b$,$a, {A_1}, {A_2}, b$ are in $A.P.$Thus,${A_1} - a = b - {A_2} \Rightarrow {A_1} + {A_2} = a + b$ ......$(i)$Since ${G_1}, {G_2}$ are $GMs$ between $a$ and $b$,$a, {G_1}, {G_2}, b$ are in $G.P.$Thus,$\frac{{G_1}}{a} = \frac{b}{{G_2}} \Rightarrow {G_1}{G_2} = ab$ ......$(ii)$Since ${H_1}, {H_2}$ are $HMs$ between $a$ and $b$,$a, {H_1}, {H_2}, b$ are in $H.P.$Thus,$\frac{1}{{H_1}} - \frac{1}{a} = \frac{1}{b} - \frac{1}{{H_2}} \Rightarrow \frac{1}{{H_1}} + \frac{1}{{H_2}} = \frac{1}{a} + \frac{1}{b}$$\Rightarrow \frac{{H_1 + H_2}}{{H_1 H_2}} = \frac{{a + b}}{{ab}}$Substituting $(i)$ and $(ii)$ into the expression,we get $\frac{{H_1 + H_2}}{{H_1 H_2}} = \frac{{A_1 + A_2}}{{G_1 G_2}}$Therefore,$\frac{{G_1 G_2}}{{H_1 H_2}} = \frac{{A_1 + A_2}}{{H_1 + H_2}}$.

If ${A_1}, {A_2}$; ${G_1}, {G_2}$ and ${H_1}, {H_2}$ are $AMs$,$GMs$,and $HMs$ between two quantities,then the value of $\frac{{G_1 G_2}}{{H_1 H_2}}$ is

Similar Questions

If the $7^{th}$ term of a $H.P.$ is $\frac{1}{10}$ and the $12^{th}$ term is $\frac{1}{25}$,then the $20^{th}$ term is

The arithmetic mean of the nine numbers in the given set $\{9, 99, 999, \dots, 999999999\}$ is a $9$-digit number $N$,all whose digits are distinct. The number $N$ does not contain the digit

Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3$. If $\sum_{i=1}^{n} f(i)=363$,then $n$ is equal to

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

The sum of the first and third term of an arithmetic progression is $12$ and the product of the first and second term is $24$. Find the first term.

Vedclass Test Series

Exam Paper Generator

Online Exam Module