Question

Let ${\alpha }_1,{\alpha }_2$ be the roots of $x^2-x+p=0$ and ${\alpha }_3,{\alpha }_4$ are the roots of $x^2-4x+q=0$ if ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are in $G.P$. Then, the integral value of $p$ and $q$ are

• A. $-2,-32$
• B. $-2,3$
• C. $-6,32$
• D. $-6,23$

Answer 1

The correct option is A $-2,-32$

Let the terms of the G.P be
$a,ar,a{r}^2,a{r}^3$
Where $a$ is the first term, and $r$ is the common ratio.
Applying the given conditions, we get
$a+ar=-(-1)=1$
Hence,
$a(1+r)=1$
And
$a{r}^2+a{r}^3=4$
$a{r}^2(1+r)=4$
$r^2\left[a\right(1+r\left)\right]=4$ Now since, $a(1+r)=1$
Hence,
$r^2=4$
$r=\pm 2$
Hence, if $r=2$,
$a=\frac{1}{3}$
Therefore,
$p=a.ar$
$=\frac{2}{9}$
$q=a^2.r^5$
$=\frac{32}{9}$
Similarly, if $r=-2$
Then $a=-1$
$p=-2$
$q=-32$