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[a c(c-a)+b\left(c^2-a^2\right)\right]=0 \Rightarrow(a-b)(c-a)\left(a c+2 b^2\right)=0 \text { (Using (1)) } $
$\Rightarrow a=b \text { or } a=c \text { or } 2 b^2+a c=0$
As $a , b , c$ are distinct, so $2 b ^2=- ac \Rightarrow a , b , \frac{- c }{2}$ are in G. Ans.
Alternatively for S-2: $b^2=\frac{2 a^2 c^2}{a^2+c^2}$
$\frac{(a+c)^2}{4}=\frac{2 a^2 c^2}{(a+c)^2-2 a c} \Rightarrow(a+c)^4-2 a c(a+c)^2-8 a^2 c^2=0$
Let $ \frac{(a+c)^2}{a c}=t$
$t ^2-2 t -8=0 \Rightarrow ( t -4)( t +2)=0 \Rightarrow t =4 \text { or } t =-2 $
$( a + c )^2=4 ac \text { or } ( a + c )^2=-2 ac$
$4 b ^2=4 ac \text { or } 4 b ^2=-2 ac $
$\therefore b ^2= ac \text { or } \therefore b ^2=\frac{- ac }{2}$
$b , c \rightarrow \text { G.P. } \therefore a , b , \frac{- c }{2} \text { are in G.P. }$
If $a , b , c \rightarrow GP . \therefore a , b , \frac{- c }{2}$ are in G.P.
and $a , b , c \rightarrow$ A.P.
then $a = b = c$ not possible.