Question

Let p and q be the roots of the equation $x^2-2x+A=0,$ and let r and s be the roots of the equation $x^2-18x+B=0.$ If p < q < r < s are in arithmetic progression. The value of (A+B) equals

• A. $80$
• B. $77$
• C. $75$
• D. $74$

Answer 1

The correct option is D $74$

Solution :-

Given, p and q be the roots of the equation

$x^2-2x+A=0$. So,

$p+q=22.....\left(1\right)$

$pq=A...\left(2\right)$

And, r and s be the roots of the equation

$x^2-18+B=0$. So,

$r+s=\mathrm{18...}\left(3\right)$

$rs=B...\left(4\right)$

Now, $p,q,r$ and s are in A.P

so let, $p=a,q=a+d,r=a+2d,s=a+3d$

Now, put these values in equation (1) and (3), we have

$a+a+d=2\Rightarrow 2a+d=\mathrm{2...}\left(5\right)$

And,

$a+2a+a+3d=18\Rightarrow 2a+5d=\mathrm{18...}\left(6\right)$

solving equation (5) and (6),

$a=-1,$ $d=4$

so $p=-1,q=-1+4=3,r=-1+8,s=-1+12=11$

Thus,

$A=pq=-1\times 3=-3$

$B=rs=7\times 11=77$

Then $A+B=-3+77=74$