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