Question

Column I		Column II
A	If $a, b, c$ are in G.P. and $a^x=b^y=c^z$ then $x, y, z$ are in	P	A.P
B	If $a \left(\frac{1}{ b }+\frac{1}{ c }\right), b \left(\frac{1}{ c }+\frac{1}{ a }\right), c \left(\frac{1}{ a }+\frac{1}{ b }\right)$ are 3 distinct terms of an A.P. then a, b, c are in	Q	G.P
C	If $a , b , c , d$ and $p$ are different real numbers such that $\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0$ then $a , b , c , d$ are in	R	H.P
D		S	Not in A.P. / G.P. / H.P.

• A. (A) R; (B) P; (C)P
• B. (A) R; (B) P; (C) Q
• C. (A) R; (B) P; (C) R
• D. (A) R; (B) P; (C) S

Answer 1

Answer: (B)

(A)$a^x=b^y=c^z=k \text { (say) }$
$a = k ^{\frac{1}{ x }} ; b = k ^{\frac{1}{y}} ; c = k ^{\frac{1}{2}}$
Given a, b, c in G.P.
$\therefore b^2=a c \Rightarrow k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}} \Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{y} \Rightarrow x, y, z \text { are in H.P. }$
(B)$\text { Given }\left(\frac{b}{c}+\frac{b}{a}\right)-\left(\frac{a}{b}+\frac{a}{c}\right)=\left(\frac{c}{a}+\frac{c}{b}\right)-\left(\frac{b}{c}+\frac{b}{a}\right) $
$\Rightarrow \left(\frac{b}{a}-\frac{a}{b}\right)+\left(\frac{b-a}{c}\right)=\left(\frac{c}{b}-\frac{b}{c}\right)+\left(\frac{c-b}{a}\right) $
$\Rightarrow \frac{b^2-a^2}{a b}+\frac{b-a}{c}=\frac{c^2-b^2}{b c}+\frac{c-b}{a}$
$\Rightarrow (b-a)\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]=(c-b)\left[\frac{1}{c}+\frac{1}{b}+\frac{1}{a}\right]$
$\therefore b-a=c-b \Rightarrow 2 b=a+c \Rightarrow a, b \text {, c are in A.P. }$
(C)$\text { We have, }\left( a ^2+ b ^2+ c ^2\right) p ^2-2( ab + bc + cd ) p +\left( b ^2+ c ^2+ d ^2\right) \leq 0 $
$\Rightarrow \left( a ^2 p ^2-2 abp + b ^2\right)+\left( b ^2 p ^2-2 bcp + c ^2\right)+\left( c ^2 p ^2-2 cdp + d ^2\right) \leq 0 $
$\Rightarrow ( ap - b )^2+( bp - c )^2+( cp - d )^2 \leq 0$
$\therefore \text { ap }- b =0, bp - c =0 \text { and } cp - d =0 .$
$\therefore p =\frac{ b }{ a }=\frac{ c }{ b }=\frac{ d }{ c } \Rightarrow a , b , c , d \text { are in G.P. }$