Sequences And Series Ques 63
Let $\alpha, \beta$ be the roots of $x^{2}-x+p=0$ and $\gamma, \delta$ be the roots of $x^{2}-4 x+q=0$. If $\alpha, \beta, \gamma, \delta$ are in GP, then the integer values of $p$ and $q$ respectively are
(2001, 1M)
(a) $-2,-32$
(b) $-2,3$
(c) $-6,3$
(d) $-6,-32$
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Answer:
Correct Answer: 63.(a)
Solution:
Formula:
$ \alpha+\beta=1 \quad \text { and } \quad \lambda+\delta=4 $
$ \alpha \beta=p \quad \text { and } \quad \lambda \delta=q $
Let $r$ be the common ratio.
Since, $\alpha, \beta$, $\gamma$ and $\delta$ are in GP.
Therefore, $\quad \beta=\alpha r, \gamma=\alpha r^{2}$
and $\quad \delta=\alpha r^{3}$
Then, $\quad \alpha+\alpha r=1 \Rightarrow \alpha(1+r)=1$ $\quad$ …….(i)
and $\quad \alpha r^{2}+\alpha r^{3}=4 \Rightarrow \alpha r^{2}(1+r)=4$ $\quad$ …….(ii)
From Eqs. (i) and (ii), $r^{2}=4 \Rightarrow r= \pm 2$
Now, $\quad \alpha \cdot \alpha r=p$ and $\alpha r^{2} \cdot \alpha r^{3}=q$
On putting $ \quad r=-2 \text {, we get } $
$ \alpha=-1, p=-2 \text { and } q=-32 $
Again putting $r=2$, we get $\alpha=1 / 3$ and $p=-\frac{2}{9}$
Since, $q$ and $p$ are integers.
Therefore, we take $p=-2$ and $q=-32$.
BETA
