Question

Let $p$ and $q$ be the roots of the equation $x^2-2x+c=0$ and $r$ and $s$ be the roots of the equation $x^2-18x+d=0$. If $p<q<r<s$ are in A. P. and values of $c$ and $d$ are

• A. $c=-3$ , $d=77$
• B. $c=3$ , $d=77$
• C. $c=3$ , $d=7$
• D. $c=3$ , $d=-7$

Answer 1

Answer: (A)

$c=-3$ , $d=77$

Given, $p$ and $q$ be the roots of the equation $x^2-2x+c=0$
$p+q=2$ and $pq=c$ ....(1)
Given, $r$ and $s$ be the roots of the equation $x^2-18x+d=0$
$r+s=18$ and $rs=d$ ....(2)
Also, given $p,q,r,s$ are in A.P.
Let the four terms in A.P. be $a-3d,a-d,a+d,a+3d$
$\Rightarrow p=a-3d,q=a-d,r=a+d,s=a+3d$
Put these values in (1) and (2), we get
$2a-4d=2$
$2a+4d=18$
$\Rightarrow a=5,d=2$
So, $p=-1,q=3,r=7,s=11$
Hence, $c=pq=-3$ and $d=rs=77$