Question

If $a,b,c$ are in harmonic progression, then

• A. $\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$ are in H.P.
• B. $\frac{2}{b}=\frac{1}{b-a}+\frac{1}{b-c}$
• C. $a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}$ are in G.P.
• D. $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in H.P.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$ are in H.P.
B $\frac{2}{b}=\frac{1}{b-a}+\frac{1}{b-c}$
C $a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}$ are in G.P.
D $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in H.P.

(A) Given, $a,b,c$ are in H.P.
$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
$\Rightarrow \frac{a+b+c}{a},\frac{b+c+a}{b},\frac{c+a+b}{c}$ are in A.P.
$\Rightarrow \frac{a+b+c}{a}-2,\frac{b+c+a}{b}-2,\frac{c+a+b}{c}-2$ are in A.P.
$\Rightarrow \frac{a+b+c-2a}{a},\frac{b+c+a-2b}{b},\frac{c+a+b-2c}{c}$ are in A.P.
$\Rightarrow \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P.
Thus, $\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$ are in H.P.


(B) Since $a,b,c$ are in H.P.,
So, the harmonic mean:
$b=\frac{2ac}{a+c}$
$a+c=\frac{2ac}{b}$ .......(i)
Cosider $\frac{1}{b-a}+\frac{1}{b-c}=\frac{b-c+b-a}{(b-a)(b-c)}$
$=\frac{2b-(a+c)}{(b-a)(b-c)}$
$=\frac{2b-(a+c)}{b^2-b(a+c)+ac}$
Substitute the value of equation(i)
$=\frac{2b-\frac{2ac}{b}}{b^2-b\left(\frac{2ac}{b}\right)+ac}$
$=\frac{2}{b}\frac{b^2-ac}{b^2-ac}$
$=\frac{2}{b}$
Therefore, $\frac{1}{b-a}+\frac{1}{b-c}=\frac{2}{b}$


(C) Let us verify geometric mean for given three numbers, then consider
$(a-\frac{b}{2})\times (c-\frac{b}{2})$
$=ac-\frac{b}{2}(a+c)+\frac{b^4}{4}$
$=ac-\frac{b}{2}\frac{2ac}{b}+\frac{b^2}{4}$ [ from equation (i)]
$=\frac{b^2}{4}$
$=(\frac{b}{2}{)}^2$
$\therefore a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}$ are in G.P.


(D) Since, we have $\frac{a+b+c}{a},\frac{b+c+a}{b},\frac{c+a+b}{c}$ are in A.P.
$\frac{a+b+c}{a}-1,\frac{b+c+a}{b}-1,\frac{c+a+b}{c}-1$ in A.P.
$\Rightarrow \frac{a+b+c-a}{a},\frac{b+c+a-b}{b},\frac{c+a+b-c}{c}$ in A.P
$\Rightarrow \frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ in A.P
$\therefore \frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ in H.P.