Question

Match the following by appropriately matching the lists based on the information given in Column I and Column II​​​​​​
$\begin{array}{cccc}& ColumnI& & ColumnII\\ a.& \text{If}a,b,c\text{are in}G.P.,\text{then}& \text{p.}& A.P.\\ & {\log }_{a}10,{\log }_{b}10,{\log }_{c}10\text{are in}& & \\ b.& \text{If}\frac{a+b{e}^x}{a-b{e}^x}=\frac{b+c{e}^x}{b-c{e}^x}=\frac{c+d{e}^x}{c-d{e}^x},& \text{q.}& H.P.\\ & \text{then}a,b,c,d& & \\ c.& \text{If}a,b,c\text{are in}A.P.;& \text{r.}& G.P.\\ & a,x,b\text{are in}G.P.\text{and}& & \\ & b,y,c\text{are in}G.P.,& & \\ & \text{then}{x}^2,b^2,y^2\text{are in}& & \end{array}$

• A. $a\to p,b\to r,c\to q$
• B. $a\to q,b\to p,c\to r$
• C. $a\to q,b\to r,c\to p$
• D. $a\to r,b\to q,c\to p$

Answer 1

The correct option is C $a\to q,b\to r,c\to p$

$a.$
$a,b,c$ are in G.P.
$b^2=ac\Rightarrow 2{\log }_{10}b={\log }_{10}a+{\log }_{10}c\Rightarrow \frac{2}{{\log }_{b}10}=\frac{1}{{\log }_{a}10}+\frac{1}{{\log }_{c}10}$
So,
${\log }_{a}10,{\log }_{b}10,{\log }_{c}10$ are in H.P.
$\left(a\right)\to \left(q\right)$

$b.$
$\frac{a+b{e}^x}{a-b{e}^x}=\frac{b+c{e}^x}{b-c{e}^x}=\frac{c+d{e}^x}{c-d{e}^x}\Rightarrow \frac{a+b{e}^x}{a-b{e}^x}+1=\frac{b+c{e}^x}{b-c{e}^x}+1=\frac{c+d{e}^x}{c-d{e}^x}+1\Rightarrow \frac{2a}{a-b{e}^x}=\frac{2b}{b-c{e}^x}=\frac{2c}{c-d{e}^x}\Rightarrow \frac{a-b{e}^x}{a}=\frac{b-c{e}^x}{b}=\frac{c-d{e}^x}{c}\Rightarrow 1-\frac{b}{a}{e}^x=1-\frac{c}{b}{e}^x=1-\frac{d}{c}{e}^x\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
Hence $a,b,c,d$ are in G.P.
$\left(b\right)\to \left(r\right)$

$c.$
$2b=a+c{x}^2=ab{y}^2=bc\Rightarrow x^2+y^2=b(a+c)=2b^2\Rightarrow b^2=\frac{x^2+y^2}{2}$
Hence $x^2,b^2,y^2$ are in A.P.
$\left(c\right)\to \left(p\right)$