Geometric progression Questions in English

(D) Let the three numbers in $G.P.$ be $a, ar, ar^2$,where $a, ar, ar^2 \in \{1, 2, \ldots, 100\}$.
Since $a, ar, ar^2$ are integers,$r$ must be a rational number. Let $r = \frac{p}{q}$ in its simplest form,where $gcd(p, q) = 1$ and $p > q \geq 1$.
The terms are $a, a\frac{p}{q}, a\frac{p^2}{q^2}$. For these to be integers,$a$ must be a multiple of $q^2$. Let $a = k q^2$.
The terms are $k q^2, k q p, k p^2$. Since $k p^2 \leq 100$,we have $k \leq \frac{100}{p^2}$.
We iterate over possible values of $p$ and $q$ such that $p > q \geq 1$ and $p^2 \leq 100$:
For $p=2, q=1$: $k \leq \frac{100}{4} = 25$. ($25$ ways)
For $p=3, q=1$: $k \leq \frac{100}{9} = 11$. ($11$ ways)
For $p=3, q=2$: $k \leq \frac{100}{9} = 11$. ($11$ ways)
For $p=4, q=1$: $k \leq \frac{100}{16} = 6$. ($6$ ways)
For $p=4, q=3$: $k \leq \frac{100}{16} = 6$. ($6$ ways)
For $p=5, q=1$: $k \leq \frac{100}{25} = 4$. ($4$ ways)
For $p=5, q=2$: $k \leq \frac{100}{25} = 4$. ($4$ ways)
For $p=5, q=3$: $k \leq \frac{100}{25} = 4$. ($4$ ways)
For $p=5, q=4$: $k \leq \frac{100}{25} = 4$. ($4$ ways)
For $p=6, q=1$: $k \leq \frac{100}{36} = 2$. ($2$ ways)
For $p=6, q=5$: $k \leq \frac{100}{36} = 2$. ($2$ ways)
For $p=7, q=1$: $k \leq \frac{100}{49} = 2$. ($2$ ways)
For $p=7, q=2, 3, 4, 5, 6$: $k \leq \frac{100}{49} = 2$. ($5 \times 2 = 10$ ways)
For $p=8, q=1, 3, 5, 7$: $k \leq \frac{100}{64} = 1$. ($4$ ways)
For $p=9, q=1, 2, 4, 5, 7, 8$: $k \leq \frac{100}{81} = 1$. ($6$ ways)
For $p=10, q=1, 3, 7, 9$: $k \leq \frac{100}{100} = 1$. ($4$ ways)
Summing these up correctly for distinct $G.P.$ sequences,the total count is $53$.

(C) Since $a, b$ and $c$ form a geometric progression,we have $b = ar$ and $c = ar^2$.
Substituting these into the line equation $ax + by + c = 0$,we get $ax + ary + ar^2 = 0$.
Dividing by $a$ (assuming $a \neq 0$),we get $x + ry + r^2 = 0$,which implies $x = -ry - r^2$.
Substituting $x = -ry - r^2$ into the curve equation $x + 2y^2 = 0$,we get $-ry - r^2 + 2y^2 = 0$,or $2y^2 - ry - r^2 = 0$.
This is a quadratic equation in $y$ of the form $Ay^2 + By + C = 0$,where $A = 2, B = -r, C = -r^2$.
The sum of the ordinates (roots of the quadratic equation) is given by $-\frac{B}{A} = -\frac{-r}{2} = \frac{r}{2}$.

(A) The given series is $1 - \frac{2}{3} + \frac{2 \cdot 4}{3 \cdot 6} - \frac{2 \cdot 4 \cdot 6}{3 \cdot 6 \cdot 9} + \ldots \infty$.
This can be rewritten as $1 - \frac{2}{3} + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^3 + \ldots \infty$.
This is an infinite geometric series with first term $a = 1$ and common ratio $r = -\frac{2}{3}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$.
Substituting the values,$S = \frac{1}{1 - (-\frac{2}{3})} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}$.

(D) Given that $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in geometric progression $(GP)$ with the same common ratio $r$.
Let $x_1 = a, x_2 = ar, x_3 = ar^2$ and $y_1 = b, y_2 = br, y_3 = br^2$.
The points are $A(a, b), B(ar, br), C(ar^2, br^2)$.
To check if they are collinear,we calculate the area of the triangle formed by these points:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$= \frac{1}{2} |a(br - br^2) + ar(br^2 - b) + ar^2(b - br)|$
$= \frac{1}{2} |abr(1 - r) + abr(r^2 - 1) + abr^2(1 - r)|$
$= \frac{1}{2} |abr - abr^2 + abr^3 - abr + abr^2 - abr^3| = 0$.
Since the area of the triangle is $0$,the points are collinear.

(C) Let the roots be $\alpha = \frac{a}{r}, \beta = a, \gamma = ar$. Since they are in geometric progression,the product of the roots is given by $\frac{a^3}{r} \cdot r = a^3 = \frac{24}{3} = 8$.
Thus,$a = 2$.
The sum of the roots is $\frac{a}{r} + a + ar = \frac{26}{3}$.
Substituting $a = 2$,we get $\frac{2}{r} + 2 + 2r = \frac{26}{3}$.
Dividing by $2$,we get $\frac{1}{r} + 1 + r = \frac{13}{3}$,which simplifies to $\frac{1}{r} + r = \frac{10}{3}$.
Multiplying by $3r$,we get $3r^2 - 10r + 3 = 0$.
Factoring gives $(3r - 1)(r - 3) = 0$,so $r = 3$ or $r = \frac{1}{3}$.
Since $\alpha < \beta < \gamma$,we must have $r > 1$,so $r = 3$.
The roots are $\alpha = \frac{2}{3}, \beta = 2, \gamma = 6$.
Finally,$3\alpha + 2\beta + \gamma = 3(\frac{2}{3}) + 2(2) + 6 = 2 + 4 + 6 = 12$.

(A) Let the roots of the equation $3x^3-26x^2+52x-24=0$ be $\frac{a}{r}, a, ar$.
We know that for a cubic equation $Ax^3+Bx^2+Cx+D=0$,the product of the roots is $-\frac{D}{A}$.
Thus,$\frac{a}{r} \times a \times ar = -\frac{(-24)}{3} = 8$.
$a^3 = 8 \Rightarrow a = 2$.
Since $a=2$ is a root,it must satisfy the equation: $3(2)^3 - 26(2)^2 + 52(2) - 24 = 24 - 104 + 104 - 24 = 0$.
Now,the sum of the roots is $\frac{a}{r} + a + ar = -\frac{(-26)}{3} = \frac{26}{3}$.
Substituting $a=2$: $\frac{2}{r} + 2 + 2r = \frac{26}{3} \Rightarrow \frac{2}{r} + 2r = \frac{26}{3} - 2 = \frac{20}{3}$.
Dividing by $2$: $\frac{1}{r} + r = \frac{10}{3} \Rightarrow 3r^2 - 10r + 3 = 0$.
Solving for $r$: $(3r-1)(r-3) = 0$,so $r = 3$ or $r = \frac{1}{3}$.
The roots are $\frac{2}{3}, 2, 6$.
The possible sums of two roots are $\frac{2}{3}+2 = \frac{8}{3}$,$2+6 = 8$,and $\frac{2}{3}+6 = \frac{20}{3}$.
Comparing with the options,the correct sum is $\frac{8}{3}$.

(C) Given that $\alpha, \beta, \gamma$ are the roots of $ax^3+bx^2+cx+d=0$ and $\beta^2=\alpha \gamma$,it implies that $\alpha, \beta, \gamma$ are in a Geometric Progression ($G$.$P$.).
Now,consider the equation $ak^3x^3+bk^2x^2+ckx+d=0$. This can be rewritten as $a(kx)^3+b(kx)^2+c(kx)+d=0$.
Let $y = kx$. Then the equation becomes $ay^3+by^2+cy+d=0$,which has roots $\alpha, \beta, \gamma$.
Therefore,$kx = \alpha, kx = \beta, kx = \gamma$,which gives $x = \frac{\alpha}{k}, x = \frac{\beta}{k}, x = \frac{\gamma}{k}$.
Thus,the roots $u, v, w$ are $\frac{\alpha}{k}, \frac{\beta}{k}, \frac{\gamma}{k}$.
Since $\alpha, \beta, \gamma$ are in $G$.$P$.,their scaled values $\frac{\alpha}{k}, \frac{\beta}{k}, \frac{\gamma}{k}$ are also in $G$.$P$.
Therefore,$u, v, w$ are in $G$.$P$.,which implies $v^2=uw$.

(D) Given the cubic equation: $x^3-14x^2+56x-64=0$.
By Vieta's formulas,for roots $\alpha, \beta, \gamma$:
$\alpha+\beta+\gamma=14$
$\alpha\beta+\beta\gamma+\gamma\alpha=56$
$\alpha\beta\gamma=64$
By testing small integer values,we find that $x=2$ is a root:
$2^3-14(2^2)+56(2)-64 = 8-56+112-64 = 0$.
Dividing the polynomial by $(x-2)$,we get:
$(x-2)(x^2-12x+32)=0$
$(x-2)(x-4)(x-8)=0$
The roots are $\alpha=2, \beta=4, \gamma=8$.
Since $\frac{4}{2} = \frac{8}{4} = 2$,the roots are in Geometric Progression $(GP)$.

(D) Given the equation: $x^3-14x^2+56x-64=0$.
Testing for roots,at $x=2$: $2^3-14(2^2)+56(2)-64 = 8-56+112-64 = 0$.
So,$(x-2)$ is a factor.
Dividing the polynomial by $(x-2)$,we get $x^2-12x+32=0$.
Factoring the quadratic: $(x-4)(x-8)=0$.
The roots are $2, 4, 8$.
Since $\frac{4}{2} = 2$ and $\frac{8}{4} = 2$,the common ratio is $2$.
Therefore,the roots are in $GP$.

(A) Let the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ be $\frac{a}{r}, a, ar$.
From the relation between roots and coefficients:
Sum of roots: $\frac{a}{r}+a+ar = -\frac{6p}{8} = -\frac{3p}{4} \Rightarrow a(\frac{1}{r}+1+r) = -\frac{3p}{4} \dots (i)$
Sum of roots taken two at a time: $(\frac{a}{r} \cdot a) + (a \cdot ar) + (\frac{a}{r} \cdot ar) = \frac{3q}{8}$ $\Rightarrow a^2(\frac{1}{r}+r+1) = \frac{3q}{8} \dots (ii)$
Product of roots: $\frac{a}{r} \cdot a \cdot ar = -(\frac{-27}{8}) = \frac{27}{8}$ $\Rightarrow a^3 = \frac{27}{8}$ $\Rightarrow a = \frac{3}{2} \dots (iii)$
Dividing (ii) by $(i)$: $\frac{a^2(\frac{1}{r}+r+1)}{a(\frac{1}{r}+r+1)} = \frac{3q/8}{-3p/4}$ $\Rightarrow a = \frac{3q}{8} \cdot (-\frac{4}{3p}) = -\frac{q}{2p} \dots (iv)$
Equating (iii) and (iv): $\frac{3}{2} = -\frac{q}{2p} \Rightarrow q = -3p$.
Now,substitute $q = -3p$ into the expression $q^2+9p^2+6pq+\frac{q}{p}$:
$(-3p)^2 + 9p^2 + 6p(-3p) + \frac{-3p}{p} = 9p^2 + 9p^2 - 18p^2 - 3 = 18p^2 - 18p^2 - 3 = -3$.

(C) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \cos x$.
For the sum to exist,$|\cos x| < 1$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
So,$\frac{1}{1-\cos x} = 4+2 \sqrt{3}$.
$1-\cos x = \frac{1}{4+2 \sqrt{3}} = \frac{4-2 \sqrt{3}}{(4+2 \sqrt{3})(4-2 \sqrt{3})} = \frac{4-2 \sqrt{3}}{16-12} = \frac{4-2 \sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.
Therefore,$\cos x = \frac{\sqrt{3}}{2}$.
The general solution for $\cos x = \cos \theta$ is $x = 2n\pi \pm \theta$.
Here,$\cos x = \cos \frac{\pi}{6}$,so $x = 2n\pi \pm \frac{\pi}{6} = (12n \pm 1) \frac{\pi}{6}$.

(B) Let $S_i$ be the number of students who gave wrong answers to at least $i$ questions. We are given $S_i = 2^{n-i}$.
The total number of wrong answers is the sum of the number of students who gave wrong answers to at least $i$ questions for all $i$ from $1$ to $n$.
Total wrong answers $= \sum_{i=1}^{n} S_i = \sum_{i=1}^{n} 2^{n-i}$.
This is a geometric series: $2^{n-1} + 2^{n-2} + \ldots + 2^0$.
Using the sum formula for a geometric series $\sum_{k=0}^{n-1} 2^k = \frac{2^n - 1}{2 - 1} = 2^n - 1$.
Given that the total number of wrong answers is $2047$,we have $2^n - 1 = 2047$.
$2^n = 2048$.
Since $2048 = 2^{11}$,we get $n = 11$.

(B) Let the sides of the right-angled triangle be $a, ar, ar^2$ where $ar^2$ is the hypotenuse.
By Pythagoras theorem,$(ar^2)^2 = a^2 + (ar)^2$.
Dividing by $a^2$ (assuming $a \neq 0$),we get $r^4 = 1 + r^2$,which implies $r^4 - r^2 - 1 = 0$.
Solving for $r^2$ using the quadratic formula,$r^2 = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 + \sqrt{5}}{2}$ (since $r^2 > 0$).
Thus,$r = \sqrt{\frac{\sqrt{5}+1}{2}}$.
In the triangle,$\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{ar}{a} = r = \sqrt{\frac{\sqrt{5}+1}{2}}$.
Similarly,$\tan \beta = \frac{a}{ar} = \frac{1}{r} = \frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}} = \sqrt{\frac{2}{\sqrt{5}+1}} = \sqrt{\frac{2(\sqrt{5}-1)}{5-1}} = \sqrt{\frac{\sqrt{5}-1}{2}}$.
Therefore,the values are $\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$.

(B) The particle moves in a sequence of steps. The $x$-coordinates and $y$-coordinates change as follows:
$x$-coordinates: $1, 1, 1 + \frac{1}{4}, 1 + \frac{1}{4}, 1 + \frac{1}{4} + \frac{1}{16}, \dots$
This is a geometric series for the horizontal movements: $x_{\infty} = 1 + \frac{1}{4} + \frac{1}{16} + \dots = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
$y$-coordinates: $0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} - \frac{1}{8}, \frac{1}{2} - \frac{1}{8}, \dots$
This is a geometric series for the vertical movements: $y_{\infty} = \frac{1}{2} - \frac{1}{8} + \frac{1}{32} - \dots = \frac{1/2}{1 - (-1/4)} = \frac{1/2}{5/4} = \frac{2}{5}$.
Thus,$(\alpha, \beta) = (\frac{4}{3}, \frac{2}{5})$.

(B) Let $a_{n}$ be the general term of a $GP$ whose first term is $a$ and common ratio is $r$.
According to the problem,each term is equal to the sum of the next two terms:
$a_{n} = a_{n+1} + a_{n+2}$
$a r^{n-1} = a r^{n} + a r^{n+1}$
Since the terms are positive,$a \neq 0$ and $r > 0$. Dividing by $a r^{n-1}$:
$1 = r + r^{2}$
$r^{2} + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^{2} - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$
Since the $GP$ consists of positive terms,the common ratio $r$ must be positive.
Therefore,$r = \frac{\sqrt{5}-1}{2}$.

(C) Let the six terms of the $GP$ be $\frac{a}{r^5}, \frac{a}{r^3}, \frac{a}{r}, ar, ar^3, ar^5$.
Given that the product of these terms is $1000$:
$\frac{a}{r^5} \cdot \frac{a}{r^3} \cdot \frac{a}{r} \cdot ar \cdot ar^3 \cdot ar^5 = 1000$
$a^6 = 1000 = 10^3$
$a^2 = (10^3)^{1/3} = 10$.
Given the fourth term is $1$:
$ar = 1 \Rightarrow a^2r^2 = 1$.
Substituting $a^2 = 10$:
$10r^2 = 1 \Rightarrow r^2 = \frac{1}{10}$.
The last term is $ar^5 = \sqrt{a^2(r^2)^5} = \sqrt{10 \cdot (\frac{1}{10})^5} = \sqrt{\frac{1}{10^4}} = \frac{1}{100}$.

(C) Let $a$ be the first term and $r$ be the common ratio of a $GP$. The $P^{\text{th}}$,$Q^{\text{th}}$,and $R^{\text{th}}$ terms are $ar^{P-1}$,$ar^{Q-1}$,and $ar^{R-1}$ respectively.
According to the question:
$ar^{P-1} = 64$ $(i)$
$ar^{Q-1} = 27$ (ii)
$ar^{R-1} = 36$ (iii)
Dividing $(i)$ by (ii):
$r^{P-Q} = \frac{64}{27} = \left(\frac{4}{3}\right)^3$ (iv)
Dividing (ii) by (iii):
$r^{Q-R} = \frac{27}{36} = \frac{3}{4}$
Raising to the power of $3$:
$r^{3(Q-R)} = \left(\frac{3}{4}\right)^3 = \left(\frac{4}{3}\right)^{-3}$ $(v)$
Multiplying (iv) and $(v)$:
$r^{P-Q} \times r^{3Q-3R} = \left(\frac{4}{3}\right)^3 \times \left(\frac{4}{3}\right)^{-3} = 1$
$r^{P+2Q-3R} = r^0$
Therefore,$P+2Q-3R = 0$,which implies $P+2Q = 3R$.

(D) The sum of an infinite geometric series is given by the formula $S = \frac{a}{1-r}$,where $a$ is the first term and $r$ is the common ratio.
Given $S = \frac{4}{5}$ and $a = \frac{3}{4}$.
Substituting the values into the formula: $\frac{4}{5} = \frac{\frac{3}{4}}{1-r}$.
Multiplying both sides by $(1-r)$,we get $\frac{4}{5}(1-r) = \frac{3}{4}$.
$1-r = \frac{3}{4} \times \frac{5}{4} = \frac{15}{16}$.
$r = 1 - \frac{15}{16} = \frac{1}{16}$.

(B) Let the given expression be $S = \frac{6}{3^{26}} + \sum_{k=0}^{24} \frac{10 \cdot 2^k}{3^{25-k}}$.
This can be rewritten as $S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \sum_{k=0}^{24} \left(\frac{2}{3^{-1}}\right)^k = \frac{6}{3^{26}} + \frac{10}{3^{25}} \sum_{k=0}^{24} (6)^k$.
Using the sum formula for a geometric progression $\sum_{k=0}^{n-1} r^k = \frac{r^n - 1}{r - 1}$,we get:
$S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \left[ \frac{6^{25} - 1}{6 - 1} \right] = \frac{6}{3^{26}} + \frac{10}{3^{25}} \left[ \frac{6^{25} - 1}{5} \right]$.
$S = \frac{6}{3^{26}} + \frac{2}{3^{25}} (6^{25} - 1) = \frac{6}{3^{26}} + \frac{2 \cdot 6^{25}}{3^{25}} - \frac{2}{3^{25}}$.
Since $6^{25} = 2^{25} \cdot 3^{25}$,we have $S = \frac{6}{3^{26}} + 2 \cdot 2^{25} - \frac{2}{3^{25}} = \frac{2 \cdot 3}{3^{26}} + 2^{26} - \frac{2}{3^{25}} = \frac{2}{3^{25}} + 2^{26} - \frac{2}{3^{25}} = 2^{26}$.

(B) Let the first three terms of the $G.P.$ be $\frac{A}{r}, A, Ar$.
Given their product is $27$:
$\frac{A}{r} \cdot A \cdot Ar = 27 \implies A^3 = 27 \implies A = 3$.
The sum of the first three terms is $S = \frac{3}{r} + 3 + 3r = 3 \left( r + \frac{1}{r} + 1 \right)$.
We know that for any real $r \neq 0$,$r + \frac{1}{r} \geq 2$ or $r + \frac{1}{r} \leq -2$.
If $r + \frac{1}{r} \geq 2$,then $S \geq 3(2 + 1) = 9$.
If $r + \frac{1}{r} \leq -2$,then $S \leq 3(-2 + 1) = -3$.
Thus,the set of possible values for $S$ is $(-\infty, -3] \cup [9, \infty)$,which is $\mathbb{R} - (-3, 9)$.
Comparing this with $\mathbb{R} - (a, b)$,we get $a = -3$ and $b = 9$.
Therefore,$a^2 + b^2 = (-3)^2 + 9^2 = 9 + 81 = 90$.

(A) The sequence is a geometric progression with first term $a = 729 = 3^6$ and common ratio $r = \frac{1}{9} = 3^{-2}$.
$P_n$ is the product of the first $n$ terms: $P_n = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{\frac{n(n-1)}{2}}$.
$P_n = (3^6)^n \cdot (3^{-2})^{\frac{n(n-1)}{2}} = 3^{6n} \cdot 3^{-n(n-1)} = 3^{6n - n^2 + n} = 3^{7n - n^2}$.
Thus,$(P_n)^{\frac{1}{n}} = 3^{7-n}$.
We need to evaluate $2 \sum_{n=1}^{40} 3^{7-n} = 2(3^6 + 3^5 + \dots + 3^{7-40}) = 2(3^6 + 3^5 + \dots + 3^{-33})$.
This is a geometric series with $40$ terms,first term $A = 3^6$,and common ratio $R = \frac{1}{3}$.
Sum $= 2 \cdot 3^6 \left( \frac{1 - (1/3)^{40}}{1 - 1/3} \right) = 2 \cdot 3^6 \cdot \frac{3^{40}-1}{3^{40}} \cdot \frac{3}{2} = \frac{3^7(3^{40}-1)}{3^{40}} = \frac{3^{40}-1}{3^{33}}$.
Comparing with $\frac{3^{\alpha}-1}{3^{\beta}}$,we get $\alpha = 40$ and $\beta = 33$.
Since $\gcd(40, 33) = 1$,the value of $\alpha + \beta = 40 + 33 = 73$.

(A) The given sequence is a $G$.$P$. with common ratio $\frac{1}{\sqrt{2}}$.
Thus,$\frac{a_2/2}{a_1} = \frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}}$.
This implies $\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \frac{1}{\sqrt{2}}$,so $\frac{a_2}{a_1} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Therefore,$a_1, a_2, \ldots, a_{10}$ is a $G$.$P$. with first term $a_1$ and common ratio $R = \sqrt{2}$.
The sum of the first $10$ terms is $S_{10} = a_1 \frac{R^{10}-1}{R-1} = 62$.
$a_1 \frac{(\sqrt{2})^{10}-1}{\sqrt{2}-1} = 62$.
$a_1 \frac{2^5-1}{\sqrt{2}-1} = 62$.
$a_1 \frac{31}{\sqrt{2}-1} = 62$.
$a_1 = \frac{62(\sqrt{2}-1)}{31} = 2(\sqrt{2}-1)$.

(B) Let the $G$.$P$. be $a, ar, ar^2, \dots$. Given $a_2 \cdot a_3 \cdot a_4 = 64$,we have $(ar)(ar^2)(ar^3) = 64$,which implies $a^3 r^6 = 64$,so $ar^2 = 4$.
Given $a_1 + a_3 + a_5 = \frac{813}{7}$,we have $a + ar^2 + ar^4 = \frac{813}{7}$.
Substituting $a = \frac{4}{r^2}$,we get $\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$.
Let $x = r^2$. Then $\frac{4}{x} + 4 + 4x = \frac{813}{7}$ $\Rightarrow 4 + 4x + 4x^2 = \frac{813}{7}x$ $\Rightarrow 28 + 28x + 28x^2 = 813x$.
$28x^2 - 785x + 28 = 0$. Solving this quadratic,$x = \frac{785 \pm \sqrt{785^2 - 4(28)(28)}}{2(28)} = \frac{785 \pm 783}{56}$.
Since the terms are increasing,$r > 1$,so $x = r^2 = \frac{1568}{56} = 28$.
We need $a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6 = ar^2(1 + r^2 + r^4) = 4(1 + 28 + 28^2) = 4(1 + 28 + 784) = 4(813) = 3252$.

(C) The product of the roots $P$ of the quadratic equation $ax^2 + bx + c = 0$ is given by $P = c/a$.
Here,$P = \frac{n^2 - 2n + 2}{n^2 - 2n + 2} = 1$. Since the product is constant,its minimum value $\alpha = 1$.
The sum of the roots $S$ is given by $S = -b/a = \frac{3}{n^2 - 2n + 2}$.
To maximize $S$,we must minimize the denominator $n^2 - 2n + 2 = (n-1)^2 + 1$.
The minimum value of the denominator is $1$ (at $n=1$),so the maximum value of the sum is $\beta = 3/1 = 3$.
For the $G$.$P$.,the first term $a = \alpha = 1$ and the common ratio $r = \alpha/\beta = 1/3$.
The sum of the first $n$ terms of a $G$.$P$. is $S_n = \frac{a(1-r^n)}{1-r}$.
For $n=6$,$S_6 = \frac{1(1-(1/3)^6)}{1-1/3} = \frac{1 - 1/729}{2/3} = \frac{728/729}{2/3} = \frac{728}{729} \times \frac{3}{2} = \frac{364}{243}$.

(NONE) The minimum value of $f(\theta) = \alpha \tan^2 \theta + \beta \cot^2 \theta$ is found using the $AM$-$GM$ inequality: $f(\theta) \ge 2\sqrt{\alpha \tan^2 \theta \cdot \beta \cot^2 \theta} = 2\sqrt{\alpha\beta}$.
Since $\alpha > \beta > 0$,the maximum value of $g(\theta) = \alpha \sin^2 \theta + \beta \cos^2 \theta$ is $\alpha$.
Given $\min f(\theta) = \max g(\theta)$,we have $2\sqrt{\alpha\beta} = \alpha$.
Squaring both sides,$4\alpha\beta = \alpha^2$,which implies $\alpha = 4\beta$ (since $\alpha \neq 0$).
The first term $a = \frac{\alpha}{2\beta} = \frac{4\beta}{2\beta} = 2$.
The common ratio $r = \frac{2\beta}{\alpha} = \frac{2\beta}{4\beta} = \frac{1}{2}$.
The sum of the first $10$ terms of the $G$.$P$. is $S_{10} = a \frac{1-r^{10}}{1-r} = 2 \frac{1-(1/2)^{10}}{1-1/2} = 4(1 - \frac{1}{1024}) = 4 \cdot \frac{1023}{1024} = \frac{1023}{256}$.
Thus,$m = 1023$ and $n = 256$. Since $\gcd(1023, 256) = 1$,$m+n = 1023 + 256 = 1279$.