Roots in A.P

let $p,q,r,s$ have common difference $d$ so $p,q,r,s$ can be written as $p,p+d,p+2d,p+3d$ now using vieta's formulas we get $-2=p+p+d$ and $-18=p+2d+p+3d$ after eliminating these two equations just plug in the values of $p,d$ so we get AP $-1,3,7,11$ since $A=pq$ and $B=rs$ so using the above AP and values of $p,q,r,s$ we get the answer $\boxed{74}$

Using Vieta's formulas, we get:

$p+q=2\\ r+s=18$

Using the formula for sum in an arithmetic series;

${ S }_{ n }=\frac { n }{ 2 } (2a+(n+1)d)$

We get two linear equations;

$\frac { 2 }{ 2 } (2p+(2-1)d)=2\\ 2p+d=2\quad \longrightarrow \quad (1)\\$ ${ S }_{ 4 }-{ S }_{ 2 }=18\\ \frac { 4 }{ 2 } (2p+(4-1)d)-2=18\\ 2p+3d=10\quad \longrightarrow \quad (2)$

Solving them simultaneously gives us

$p=-1\\ d=4$

Which is enough to complete the sequence;

$....\quad p,q,r,s\quad ....\\ ....\quad -1,3,7,11\quad ....$

Now return to Vieta's formula again to get

$p.q=A\\ r.s=B$

Compute to get answer;

$A=-3\\ B=77\\ \therefore \boxed { A+B=74 }$

Given p,q,r,s is an AP. Let d be the common difference. Now consider p+q, r+s as another AP,. Now as per properties of AP, the new common difference will be four times the before, that is 4d. By writing roots of quadratic equations we will get p+q, p-q, r+s, r-s, and by applying properties of AP, we can find out the A+B value....