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