SAT1000 - P461

Given that $a_n = a_1q^{n-1}$ , where $q=\sqrt 2$ , and $\displaystyle S_n = \sum_{k=1}^n a_k = \frac {a_1(q^n-1)}{q-1}$ , implies that:

$\begin{aligned} T_n & = \frac {17S_n - S_{2n}}{a_{n+1}} \\ & = \frac {17a_1(q^n-1)-a_1(q^{2n}-1)}{a_1q^n(q-1)} \\ & = \frac {q^{2n}-17q^n+16}{q^n(1-q)} \\ & = \frac 1{1-q} \left(q^n-17+\frac {16}{q^n} \right) & \small \blue{\text{To find }\max(T_n)} \\ \frac {dT_n}{dn} & = \frac 1{1-q} \left(q^n\ln q - \frac {16 \ln q}{q^n} \right) & \small \blue{\text{Putting }\frac {dT_n}{dn} = 0} \\ q^{2n} & = 16 \\ 2^n & = 16 \\ \implies n & = 4 \end{aligned}$

Since $\dfrac {d^2T_n}{dn^2} \bigg|_{n=4} < 0$ , $\implies \max (T_n) = T_4$ . Therefore $m= \boxed 4$ .

$T_m=\dfrac {17-16\times 2^{-\frac{m}{2}}-2^{\frac{m}{2}}}{\sqrt 2 -1}$ .

This will be a maximum when $2^m=16=2^4\implies m=\boxed 4$ .

Solution to this problem :

Solving the equation $x^2-7x+16-r^2=0,$ we get $x_A=x_B=\dfrac {7-\sqrt {4r^2-15}}{2},x_C=x_D=\dfrac {7+\sqrt {4r^2-15}}{2}$

Area of the quadrilateral $ABCD$ is $\sqrt {\frac{4r^2-15}{2}}\left (\sqrt {7+\sqrt {4r^2-15}}+\sqrt {7-\sqrt {4r^2-15}}\right )$

This attains a maximum when $r^2=14.63889$

The value of $x_0=\sqrt {16-r^2}\approx 1.166666$

Therefore the required answer is $\boxed {1166}$