Question

Statement-1: Let $m,n,a,b,c$ be non-zero real numbers such that $a,b,c$ are in harmonic progression, then $\frac{a}{m+na},\frac{b}{m+nb},\frac{c}{m+nc}$ are also in harmonic progression.
Statement-2 : If a, b, c are non-zero distinct real numbers such that $a,b,c$ are in arithmetic progression and $a^2,b^2,c^2$ are in harmonic progression, then $a,b,\frac{-c}{2}$ are in geometric progression.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true

Answer 1

Answer: (B)

Given that $a,b,c$ are in H.P.,
so, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P. $\Rightarrow \frac{m}{a},\frac{m}{b},\frac{m}{c}$ are in A.P. $\Rightarrow \frac{m}{a}+n,\frac{m}{b}+n,\frac{m}{c}+n$ are in A.P.
$\Rightarrow \frac{m+na}{a},\frac{m+nb}{b},\frac{m+nc}{c}$ are in A.P. $\Rightarrow \frac{a}{m+na},\frac{b}{m+nb},\frac{c}{m+nc}$ are in H.P.
S-2 :
Since $a,b,c$ are in H.P., we have $b-a=c-b$
Also, $a^2,b^2,c^2$ are in H.P., we get
$\frac{1}{b^2}-\frac{1}{a^2}=\frac{1}{c^2}-\frac{1}{b^2}\Rightarrow \frac{a^2-b^2}{a^2{b}^2}=\frac{b^2-c^2}{b^2{c}^2}\Rightarrow (a-b)\left[c^2(a+b)-a^2(b+c)\right]=0(\text{ Using (1)) }$
$\Rightarrow (a-b)\left[ac(c-a)+b\left(c^2-a^2\right)\right]=0\Rightarrow (a-b)(c-a)\left(ac+2b^2\right)=0\text{ (Using (1)) }$
$\Rightarrow a=b\text{ or }a=c\text{ or }2b^2+ac=0$
As $a,b,c$ are distinct, so $2b^2=-ac\Rightarrow a,b,\frac{-c}{2}$ are in G. Ans.
Alternatively for S-2: $b^2=\frac{2a^2{c}^2}{a^2+c^2}$
$\frac{(a+c{)}^2}{4}=\frac{2a^2{c}^2}{(a+c{)}^2-2ac}\Rightarrow (a+c{)}^4-2ac(a+c{)}^2-8a^2{c}^2=0$
Let $\frac{(a+c{)}^2}{ac}=t$
$t^2-2t-8=0\Rightarrow (t-4)(t+2)=0\Rightarrow t=4\text{ or }t=-2$
$(a+c{)}^2=4ac\text{ or }(a+c{)}^2=-2ac$
$4b^2=4ac\text{ or }4b^2=-2ac$
$\therefore b^2=ac\text{ or }\therefore b^2=\frac{-ac}{2}$
$b,c\to \text{ G.P. }\therefore a,b,\frac{-c}{2}\text{ are in G.P. }$
If $a,b,c\to GP.\therefore a,b,\frac{-c}{2}$ are in G.P.
and $a,b,c\to$ A.P.
then $a=b=c$ not possible.