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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If the p and q are roots of the equation $x^2-7x+A=0\;\xi\;r\;\xi\;s$ are roots of equation $x^2-19x+B=0,$ find the values of $A\;\xi\;B$ if p,q,r,s are in AP.

$(a)\;A=rs,B=pq\qquad(b)\;A=B=0\qquad(c)\;A=7,B=19\qquad(d)\;A=10,B=88$

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Answer : (d) A=10 , B=88
Explanation : Equations $\;x^2-7x+A=0\qquad\;x^2-19x+B=0$
$p+q=7\qquad\;pq=A$
$r+s=7\qquad\;rs=B$
$Since\; p,q,r,s \;are\;in\;AP$
$let\;p=a-3d\qquad\;q=a-d$
$r=a+d\qquad\;,s=a+3d$
$p+q=a-3d+a-d=7=2a-4d$
$r+s=a+d+a+3d=19=2a+4d$
$Adding\;both\;equations\;\quad\;26=4a$
$a=\frac{26}{4}=\frac{13}{2}$
$2a-4d=7$
$13-4d=7\quad\;d=\frac{6}{4}=\frac{3}{2}$
$p=a-3d=\frac{13}{2}-\frac{9}{2}=\frac{4}{2}=2$
$q=a-d=\frac{13}{2}-\frac{3}{2}=5$
$r=8\quad\;s=11$
$AP\; is\; 2,5,8,11.$
$A=pq=2*5=10$
$B=rs=8*11=88.$

 

answered Jan 2, 2014 by yamini.v
edited Jan 2, 2014 by yamini.v
 

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