Let p and q be roots of the equation x^2-2x+A | vedclass

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(C) Let the four numbers in $A.P.$ be $p=a-3d, q=a-d, r=a+d, s=a+3d$. Since $p$ and $q$ are roots of $x^2-2x+A=0$,we have $p+q=2$ and $pq=A$. Since $r

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$-3, 77$

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(C) Let the four numbers in $A.P.$ be $p=a-3d, q=a-d, r=a+d, s=a+3d$. Since $p$ and $q$ are roots of $x^2-2x+A=0$,we have $p+q=2$ and $pq=A$. Since $r$ and $s$ are roots of $x^2-18x+B=0$,we have $r+s=18$ and $rs=B$. Summing the roots: $(a-3d)+(a-d)+(a+d)+(a+3d) = 4a = 2+18 = 20$,so $a=5$. From $p+q=2$,we get $(a-3d)+(a-d) = 2a-4d = 2$. Substituting $a=5$,we get $10-4d=2$,so $4d=8$,which means $d=2$. The roots are $p=5-3(2)=-1$,$q=5-2=3$,$r=5+2=7$,and $s=5+3(2)=11$. Thus,$A = pq = (-1)(3) = -3$ and $B = rs = (7)(11) = 77$.

Let $p$ and $q$ be 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 $A.P.$,then $A$ and $B$ are

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