Let $p$ and $q$ be the roots of the equation $\displaystyle x^{2} - 2x + c = 0$ and $r$ and $s$ be the roots of the equation $\displaystyle x^{2} - 18x + d = 0$. If $\displaystyle p < q < r < s$ are in A. P. and values of $c$ and $d$ are

Let the roots be p, q, r, s in A.P. with common difference d'. Then p=a, q=a+d', r=a+2d', s=a+3d'. From the first equation, p+q = 2 and pq = c. From the second, r+s = 18 and rs = d. Thus, 2a+d' = 2 and 2a+5d' = 18. Solving gives 4d' = 16, so d'=4 and a=-1. The roots are -1, 3, 7, 11. Then c = (-1)(3) = -3 and d = (7)(11) = 77.