Question

If $a,b,c$ be the $p^{th}$ , $q^{th}$ and $r^{th}$ terms respectively of an A.R and GR both, then the product of the roots of equation $(a^b{b}^c{c}^a)x^2-(abc)x+(a^c{b}^a{c}^b)=0$ equals

• A. $-1$
• B. $2$
• C. $abc$
• D. $1$

Answer 1

Answer: (D)

Product of the roots of given equation
$=\frac{a^c{b}^a{c}^b}{a^b{b}^c{c}^a}=a^{c-b}{b}^{a-c}{c}^{b-a}$
Also, $a=A+\left(p-1\right)d\}$
$b=A+(q-1)d\}$
$c=A+(r-1)d\}\dots (\ast )$ A = first term of A.P. d = common difference of A.P
and $a=B{R}^{p-1}\}$
$b=B{R}^{q-1}\}$
$c=B{R}^{r-1}\}\dots (\ast \ast )$ B is the first term and R is common ratio of GP.
Now, $a^{c-b}{b}^{a-c}{c}^{b-a}$
$={\left(B{R}^{p-1}\right)}^{\left(r-q\right)d}\times {\left(B{R}^{q-1}\right)}^{\left(p-r\right)d}\times {\left(B{R}^{r-1}\right)}^{\left(q-p\right)d}$
(From (*) and (**))
$=B^{\left(r-q+p-r+q-p\right)d}$
$R^{\left[\left(p-1\right)\left(r-q\right)+\left(q-1\right)\left(p-r\right)+\left(r-1\right)\left(q-p\right)\right]d}=B^0{R}^0=1$