Sequences and Series 2 Question 14
15. Let $p$ and $q$ be the roots of the equation $x^{2}-2 x+A=0$ and let $r$ and $s$ be the roots of the equation $x^{2}-18 x+B=0$. If $p<q<r<s$ are in arithmetic progression, then $A=\ldots$ and $B=\ldots$.
(1997, 2M)
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Answer:
Correct Answer: 15. (9)
Solution:
- Given, $p+q=2, p q=A$
and $\quad r+s=18, r s=B$
and it is given that $p, q, r, s$ are in an AP.
Therefore, let $p=a-3 d, q=a-d, r=a+d$
and
$$ s=a+3 d $$
Since, $\quad p<q<r<s$
We have, $\quad d>0$
$\begin{aligned} \text { Now, } \quad 2 & =p+q=a-3 d+a-d=2 a-4 d \ \Rightarrow \quad a-2 d & =1\end{aligned}$
Again, $\quad 18=r+s=a+d+a+3 d$
$$ 18=2 a+4 d $$
$\Rightarrow \quad 9=a+2 d$
On subtracting Eq. (i) from Eq. (ii), we get
$$ 8=4 d \Rightarrow d=2 $$
On putting in Eq. (ii), we get $a=5$
$$ \begin{aligned} \therefore \quad p & =a-3 d=5-6=-1 \\ q & =a-d=5-2=3 \\ r & =a+d=5+2=7 \end{aligned} $$
and $\quad s=a+3 d=5+6=11$
Therefore, $A=p q=-3$ and $B=r s=77$
