Question

If $G_1,G_2$ and $H_{1,}{H}_2$ be two A.Ms and H.M.s between two quantities, then the value of $\frac{G_1{G}_2}{H_1{H}_2}$ is

Answer 1

Let a and b be the two quantities .

Given,$G_1,G_2$ are two AMs between a and b.

$\therefore$ these are 4 times.

$\therefore a+(4-1)d=b.$

$\Rightarrow a+3d=b.$

$\Rightarrow 3d=b-a$

$\Rightarrow d=\frac{b-a}{3}$-----(1)

the second term is $G_1=a+d$

$=a+\frac{b-a}{3}=\frac{3a+b-a}{3}=\frac{2a+b}{3}$

$\therefore G_1=\frac{2a+b}{3}$

the third term is $G_2=a+2d$

=$a+2\left(\frac{b-a}{3}\right)=\frac{3a+2b-2a}{3}=\frac{a+2b}{3}$

$\therefore G_2=\frac{a+2b}{3}$

Thus $G_1{G}_2=\left(\frac{2a+b}{3}\right)\left(\frac{a+2b}{3}\right)$

$=\frac{(2a+b)(a-2b)}{9}$

Also, $H_1{H}_2$ are HMs Between a and b.

$\therefore a{r}^3=b$

$\Rightarrow r^3=\frac{b}{a}$

the second term is $H_1=ar.$

the third term is $H_2=a{r}^2$

$\therefore H_1{H}_2\left(ar\right)\left(a{r}^2\right)=a^2{r}^3=a^2.\frac{b}{a}=ab.$

Hence $\frac{G_1{G}_2}{H_1{H}_2}=\frac{(2a+b)(a+2b)}{\frac{9}{ab}}=\frac{(2a+b)(a+2b)}{9ab}$