Question

Match the following. If a,b,c are in HP , then

Answer 1

Since $a$, $b$ and $c$ are in H.P., then

$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

or, $\frac{2}{b}=\frac{a+c}{ac}$

or, $b=\frac{2ac}{a+c}$.......(1).

Now,

$\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$ are in H.P.

$\iff$ $\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P.

$\iff$ $\frac{b+c}{a}-1,\frac{c+a}{b}-1,\frac{a+b}{c}-1$ are in A.P.

$\iff$ $\frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b}$

$\iff$ $\frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc}$

$\iff$ $\frac{(a+b+c)(a-b)}{ab}=\frac{(a+b+c)(b-c)}{bc}$

$\iff$ $\frac{(a-b)}{ab}=\frac{(b-c)}{bc}$

$\iff \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$.

Also,

$\frac{1}{b-a},\frac{1}{b},\frac{1}{b-c}$ are in A.P.

$\iff \frac{1}{b}-\frac{1}{b-a}=\frac{1}{b-c}-\frac{1}{b}$

$\iff \frac{-a}{b(b-a)}=\frac{-c}{b(b-c)}$

$\iff \frac{(b-a)}{a}=\frac{(c-b)}{c}$

$\iff \frac{(b-a)}{ab}=\frac{(c-b)}{cb}$

$\iff \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$.

Again,

$(a-\frac{b}{2})(c-\frac{b}{2})=\frac{b^2}{4}-\frac{b}{2}(a+c)+ac=\frac{b^2}{4}$ [Using (1)].

So, $a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}$ are in G.P.

Proceeding the same way as we did for the first one it can be shown than $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in H.P.