Explanation:
A Given
\( p \text { and } q \text { are the roots of the equation } x^2\)
\(-2 x+A=0 \text {, Then }\)
\(p+q=2\)
\(p \cdot q=A\)
\(r \text { and } s \text { are the roots of the equation }\)
\(x^2-18 x+B=0\)
\(\therefore \mathrm{r}+\mathrm{s}=18\)
\(\mathrm{rs}=\mathrm{B}\)
Now, p, q,r \& \(s\) are in A.P.
Let, \(\mathrm{a}=\mathrm{p}, \mathrm{a}+\mathrm{d}=\mathrm{q}, \mathrm{a}+2 \mathrm{~d}=\mathrm{r}, \mathrm{a}+3 \mathrm{~d}=\mathrm{s}\)
Put the above value in equation (i)
\(p+q=a+a+d=2\)
\(2 a+d=2\)
And,
\(r+s=a+2 d+a+3 d=18\)
\(2 a+5 d=18\)
Solving equation (v) and (iv)-
\(\mathrm{a}=-1, \mathrm{~d}=4\)
So,
\(\text { So, } \mathrm{p}=-1, \mathrm{q}=-1+4=3\)
\(\mathrm{r}=-1+8=7, \mathrm{q}=-1+12=11\)
\(\therefore \mathrm{A}=\mathrm{pq}=-1 \times 3\)
\(\mathrm{~A}=-3\)
\(\mathrm{~B}=\mathrm{rs}\)
\(\mathrm{B}=7 \times 11=77\)