Question

If a, b, c are in A.P., $\alpha ,\beta ,\gamma$ in H.P.,$a\alpha ,b\beta ,c\gamma$ in G.P. (with common ratio not equal to 1.), then prov that a : b : c = $\frac{1}{\gamma }:\frac{1}{\beta }:\frac{1}{\alpha }$

Answer 1

2b = a + c , $\alpha =\frac{2\alpha \gamma }{\alpha +\gamma }$ and $b^2{\beta }^2=ac\alpha \gamma$
Eliminating b and $\beta$ from these , we get
$\frac{(a+c{)}^2}{4}\cdot \frac{4{\alpha }^2{\gamma }^2}{(\alpha +\gamma {)}^2}=ac\alpha \gamma$
or $\frac{(a+c{)}^2}{ac}=\frac{(\alpha +\gamma {)}^2}{\alpha \gamma }$
This gives $\frac{a}{c}+\frac{c}{a}=\frac{\alpha }{\gamma }+\frac{\gamma }{\alpha }$ Multiply by $\frac{a}{c}$
or ${\left(\frac{a}{c}\right)}^2-\frac{a}{c}(\frac{\alpha }{\gamma }+\frac{\gamma }{\alpha })+1=0$
$(\frac{a}{c}-\frac{\alpha }{\gamma })(\frac{a}{c}-\frac{\gamma }{\alpha })=0$
$\therefore \frac{a}{c}=\frac{\alpha }{\gamma }$ i.e $\frac{1}{1/\gamma }=\frac{1}{1/\alpha }$ ....(1)
or $\frac{a}{c}=\frac{\alpha }{\gamma }$ i. e. , $a\alpha =c\gamma$ ....(2)
The condition (2) is ruled out since this with the help of $b^2{\beta }^2$ = ac$\alpha \gamma$ gives b$\beta$ = c$\gamma$ so that a$\alpha$ = b$\beta$ = c$\gamma$ and so a$\alpha$ ,b$\beta$ ,c$\gamma$ are in G.P. with common ratio 1 which contradicts the hypothesis.
Now using (1) and $b^2{\alpha }^2$ = ac$\alpha \gamma$ ,we get
$b^2{\beta }^2$ = $c^2{\alpha }^2$ or $b\beta$ = $c\alpha$
Thus $\frac{b}{1/\beta }=\frac{c}{1/\alpha }$ .....(3)
Finally from (1) and (3) , we get
$\frac{a}{1/\gamma }=\frac{b}{1/\beta }=\frac{c}{1/\alpha }$