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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 AP, then find the value of A and B. (JEE advance 1997)

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    You have an unfinished sentence. Please check that! – 2017-01-14
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    Hint: $p+q=2\implies 2p+M=2$ where $M$ is the period of the progression. Similarly $r+s=18\implies 2p+5M=18$. – 2017-01-14
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    Please solve further and tell me the answer. – 2017-01-14

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Let $p,\space q,\space r,\space s$ be $(a-3d), (a-d), (a+d), (a+3d)$ respectively as they are in A.P.

Now from the first quadratic equation $$p+q=2\implies2a-4d=2\implies a-2d=1$$

From second quadratic equation $$r+s=18\implies 2a+4d=18\implies a+2d=9$$

From above equations $$a=5,\space d = 2$$

$$\therefore p = -1,\space q = 3,\space r = 7,\space s = 11$$

Hence $$A = p\cdot q = -3$$$$B = r\cdot s = 77$$