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 arithmetic progression, then A = and B =.........: Sequence & Series (Mathematics)

Let ' $a$ ' be the first term of the A.P. and ' $d$ ' the common difference. $\\$ Suppose, $p=a-3 d, q=a-d, r=a+d$ and $s=a+3 d$ $\\ \therefore p+q=2 a-4 d=2 ; r+s=2 a+4 d=18$ i.e. $\\ 2 a-4 d=2[\because p+q=2$ and $r+s=18]$ $\\ \begin{aligned} 2 a+4 d & =18 \\ 4 a & =20\end{aligned}$ $\\ \begin{aligned} & \Rightarrow a=5 \therefore d=2 \\& \Rightarrow p=a-3 d=5-3 \times 2=-1 \\& q=a-d=5-2=3 ; r=a+d=5+2=7 \\& s=a+3 d=5+6=11 \text { also } p q=A \\& \Rightarrow A=3 \times(-1)=-3\end{aligned}\\$ Also $r s=B \Rightarrow B=7 \times 11=77$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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